3.6.65 \(\int \frac {\cos ^3(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [565]

Optimal. Leaf size=241 \[ -\frac {5 a b^4 \tanh ^{-1}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}}{\left (a^2+b^2\right )^{7/2} d}+\frac {b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

-5*a*b^4*arctanh((b-a*tan(d*x+c))/(a^2+b^2)^(1/2)/(sec(d*x+c)^2)^(1/2))*cos(d*x+c)*(sec(d*x+c)^2)^(1/2)/(a^2+b
^2)^(7/2)/d+1/3*b*(2*a^4+9*a^2*b^2-8*b^4)*sec(d*x+c)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))+1/3*cos(d*x+c)^3*(b+a*tan(
d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x+c))-1/3*cos(d*x+c)*(b*(a^2-4*b^2)-a*(2*a^2+7*b^2)*tan(d*x+c))/(a^2+b^2)^2/d/(
a+b*tan(d*x+c))

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Rubi [A]
time = 0.19, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3593, 755, 837, 821, 739, 212} \begin {gather*} \frac {\cos ^3(c+d x) (a \tan (c+d x)+b)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {5 a b^4 \cos (c+d x) \sqrt {\sec ^2(c+d x)} \tanh ^{-1}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{7/2}}+\frac {b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + b*Tan[c + d*x])^2,x]

[Out]

(-5*a*b^4*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[Sec[c + d*x]^
2])/((a^2 + b^2)^(7/2)*d) + (b*(2*a^4 + 9*a^2*b^2 - 8*b^4)*Sec[c + d*x])/(3*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x
])) + (Cos[c + d*x]^3*(b + a*Tan[c + d*x]))/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]*(b*(a^2 - 4
*b^2) - a*(2*a^2 + 7*b^2)*Tan[c + d*x]))/(3*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 3593

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[d^(2*
IntPart[m/2])*((d*Sec[e + f*x])^(2*FracPart[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {-2 \left (2+\frac {a^2}{b^2}\right )-\frac {3 a x}{b^2}}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{3 \left (a^2+b^2\right ) d}\\ &=\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (b^5 \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {-\frac {2 \left (a^2-4 b^2\right )}{b^4}+\frac {a \left (2 a^2+7 b^2\right ) x}{b^6}}{(a+x)^2 \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d}\\ &=\frac {b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (5 a b^3 \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac {b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (5 a b^3 \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a^2}{b^2}-x^2} \, dx,x,\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\sec ^2(c+d x)}}\right )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac {5 a b^4 \tanh ^{-1}\left (\frac {b \left (1-\frac {a \tan (c+d x)}{b}\right )}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}}{\left (a^2+b^2\right )^{7/2} d}+\frac {b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 1.28, size = 249, normalized size = 1.03 \begin {gather*} \frac {\sec (c+d x) \left (240 a b^4 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right ) (a \cos (c+d x)+b \sin (c+d x))+\left (a^2+b^2\right ) \left (15 a^4 b+90 a^2 b^3-45 b^5+20 b^3 \left (a^2+b^2\right ) \cos (2 (c+d x))+b \left (a^2+b^2\right )^2 \cos (4 (c+d x))+10 a^5 \sin (2 (c+d x))+40 a^3 b^2 \sin (2 (c+d x))+30 a b^4 \sin (2 (c+d x))+a^5 \sin (4 (c+d x))+2 a^3 b^2 \sin (4 (c+d x))+a b^4 \sin (4 (c+d x))\right )\right )}{24 \left (a^2+b^2\right )^4 d (a+b \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + b*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]*(240*a*b^4*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c + d*x] +
b*Sin[c + d*x]) + (a^2 + b^2)*(15*a^4*b + 90*a^2*b^3 - 45*b^5 + 20*b^3*(a^2 + b^2)*Cos[2*(c + d*x)] + b*(a^2 +
b^2)^2*Cos[4*(c + d*x)] + 10*a^5*Sin[2*(c + d*x)] + 40*a^3*b^2*Sin[2*(c + d*x)] + 30*a*b^4*Sin[2*(c + d*x)] +
a^5*Sin[4*(c + d*x)] + 2*a^3*b^2*Sin[4*(c + d*x)] + a*b^4*Sin[4*(c + d*x)])))/(24*(a^2 + b^2)^4*d*(a + b*Tan[
c + d*x]))

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Maple [A]
time = 0.47, size = 320, normalized size = 1.33

method	result	size
derivativedivides	\(\frac {-\frac {2 b^{4} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {5 a \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {2 \left (\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{3} b -6 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} a^{4}-6 a^{2} b^{2}+\frac {8}{3} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 a \,b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 a^{3} b}{3}-\frac {14 a \,b^{3}}{3}\right )}{\left (a^{2}+b^{2}\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d}\)	\(320\)
default	\(\frac {-\frac {2 b^{4} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {5 a \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {2 \left (\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{3} b -6 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} a^{4}-6 a^{2} b^{2}+\frac {8}{3} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 a \,b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 a^{3} b}{3}-\frac {14 a \,b^{3}}{3}\right )}{\left (a^{2}+b^{2}\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d}\)	\(320\)
risch	\(-\frac {i {\mathrm e}^{3 i \left (d x +c \right )}}{24 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b}{8 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 b^{2} a \right ) d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a}{8 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 b^{2} a \right ) d}-\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b}{8 \left (i b +a \right )^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a}{8 \left (i b +a \right )^{3} d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{24 \left (i b +a \right )^{2} d}-\frac {2 i b^{5} {\mathrm e}^{i \left (d x +c \right )}}{\left (-i a +b \right )^{3} d \left (i a +b \right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}+\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{7}+3 i a^{5} b^{2}+3 i a^{3} b^{4}+i a \,b^{6}-a^{6} b -3 a^{4} b^{3}-3 b^{5} a^{2}-b^{7}}{\left (a^{2}+b^{2}\right )^{\frac {7}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {7}{2}} d}-\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{7}+3 i a^{5} b^{2}+3 i a^{3} b^{4}+i a \,b^{6}-a^{6} b -3 a^{4} b^{3}-3 b^{5} a^{2}-b^{7}}{\left (a^{2}+b^{2}\right )^{\frac {7}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {7}{2}} d}\)	\(449\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2*b^4/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*((-b^2/a*tan(1/2*d*x+1/2*c)-b)/(a*tan(1/2*d*x+1/2*c)^2-2*b*tan(1/2*d
*x+1/2*c)-a)-5*a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))-2/(a^2+b^2)/(a^4+2
*a^2*b^2+b^4)*((-a^4-3*a^2*b^2+2*b^4)*tan(1/2*d*x+1/2*c)^5+(-2*a^3*b-6*a*b^3)*tan(1/2*d*x+1/2*c)^4+(-2/3*a^4-6
*a^2*b^2+8/3*b^4)*tan(1/2*d*x+1/2*c)^3-8*a*b^3*tan(1/2*d*x+1/2*c)^2+(-a^4-3*a^2*b^2+2*b^4)*tan(1/2*d*x+1/2*c)-
2/3*a^3*b-14/3*a*b^3)/(1+tan(1/2*d*x+1/2*c)^2)^3)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 772 vs. \(2 (229) = 458\).
time = 0.52, size = 772, normalized size = 3.20 \begin {gather*} -\frac {\frac {15 \, a b^{4} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{5} b + 14 \, a^{3} b^{3} - 3 \, a b^{5} - \frac {15 \, a b^{5} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {{\left (3 \, a^{6} + 13 \, a^{4} b^{2} + 22 \, a^{2} b^{4} - 3 \, b^{6}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {{\left (4 \, a^{5} b + 28 \, a^{3} b^{3} - 21 \, a b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (a^{6} - 9 \, a^{4} b^{2} - 46 \, a^{2} b^{4} + 9 \, b^{6}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, {\left (2 \, a^{5} b + 6 \, a^{3} b^{3} - 5 \, a b^{5}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{6} + 3 \, a^{4} b^{2} + 38 \, a^{2} b^{4} - 9 \, b^{6}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, {\left (a^{6} + 3 \, a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} + \frac {2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(15*a*b^4*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c)
+ 1) - sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 2*(2*a^5*b + 14*a^3*b^3 - 3*
a*b^5 - 15*a*b^5*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + (3*a^6 + 13*a^4*b^2 + 22*a^2*b^4 - 3*b^6)*sin(d*x + c)/
(cos(d*x + c) + 1) + (4*a^5*b + 28*a^3*b^3 - 21*a*b^5)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - (a^6 - 9*a^4*b^2
- 46*a^2*b^4 + 9*b^6)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*(2*a^5*b + 6*a^3*b^3 - 5*a*b^5)*sin(d*x + c)^4/(
cos(d*x + c) + 1)^4 + (a^6 + 3*a^4*b^2 + 38*a^2*b^4 - 9*b^6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*(a^6 + 3*
a^4*b^2 - 2*a^2*b^4 + b^6)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + 2*(a^
7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*sin(d*x + c)/(cos(d*x + c) + 1) + 2*(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^
6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*sin(d*x + c)^3/(cos(d*x + c
) + 1)^3 + 6*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*(a^8 + 3*a^6*b^2
+ 3*a^4*b^4 + a^2*b^6)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8))/d

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Fricas [A]
time = 0.42, size = 418, normalized size = 1.73 \begin {gather*} \frac {4 \, a^{6} b + 22 \, a^{4} b^{3} + 2 \, a^{2} b^{5} - 16 \, b^{7} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b - 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (a^{2} b^{4} \cos \left (d x + c\right ) + a b^{5} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, a^{7} + 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} + 7 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} d \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(4*a^6*b + 22*a^4*b^3 + 2*a^2*b^5 - 16*b^7 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4 - 2*(a
^6*b - 2*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2 + 15*(a^2*b^4*cos(d*x + c) + a*b^5*sin(d*x + c))*sqrt(a^2
+ b^2)*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(
b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 2*((
a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (2*a^7 + 11*a^5*b^2 + 16*a^3*b^4 + 7*a*b^6)*cos(d*x + c)
)*sin(d*x + c))/((a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8)*d*cos(d*x + c) + (a^8*b + 4*a^6*b^3 + 6*a^4
*b^5 + 4*a^2*b^7 + b^9)*d*sin(d*x + c))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.60, size = 438, normalized size = 1.82 \begin {gather*} -\frac {\frac {15 \, a b^{4} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {6 \, {\left (b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{5}\right )}}{{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} - \frac {2 \, {\left (3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{3} b + 14 \, a b^{3}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(15*a*b^4*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b
+ 2*sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 6*(b^6*tan(1/2*d*x + 1/2*c) + a*
b^5)/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)) - 2*(3*
a^4*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*a^3*b*tan(1/2
*d*x + 1/2*c)^4 + 18*a*b^3*tan(1/2*d*x + 1/2*c)^4 + 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x + 1/
2*c)^3 - 8*b^4*tan(1/2*d*x + 1/2*c)^3 + 24*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) + 9*a^2*b
^2*tan(1/2*d*x + 1/2*c) - 6*b^4*tan(1/2*d*x + 1/2*c) + 2*a^3*b + 14*a*b^3)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3))/d

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Mupad [B]
time = 7.11, size = 674, normalized size = 2.80 \begin {gather*} \frac {\frac {2\,\left (2\,a^4\,b+14\,a^2\,b^3-3\,b^5\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {10\,b^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^4+6\,a^2\,b^2-5\,b^4\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^4\,b+28\,a^2\,b^3-21\,b^5\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^6+3\,a^4\,b^2-2\,a^2\,b^4+b^6\right )}{a\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^6+13\,a^4\,b^2+22\,a^2\,b^4-3\,b^6\right )}{3\,a\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^6+3\,a^4\,b^2+38\,a^2\,b^4-9\,b^6\right )}{3\,a\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^6-9\,a^4\,b^2-46\,a^2\,b^4+9\,b^6\right )}{3\,a\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {10\,a\,b^4\,\mathrm {atanh}\left (\frac {2\,a^6\,b+2\,b^7+6\,a^2\,b^5+6\,a^4\,b^3-2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}{2\,{\left (a^2+b^2\right )}^{7/2}}\right )}{d\,{\left (a^2+b^2\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + b*tan(c + d*x))^2,x)

[Out]

((2*(2*a^4*b - 3*b^5 + 14*a^2*b^3))/(3*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) - (10*b^5*tan(c/2 + (d*x)/2)^6)/(a
^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (10*b*tan(c/2 + (d*x)/2)^4*(2*a^4 - 5*b^4 + 6*a^2*b^2))/(3*(a^6 + b^6 + 3*
a^2*b^4 + 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^2*(4*a^4*b - 21*b^5 + 28*a^2*b^3))/(3*(a^2 + b^2)*(a^4 + b^4 + 2
*a^2*b^2)) - (2*tan(c/2 + (d*x)/2)^7*(a^6 + b^6 - 2*a^2*b^4 + 3*a^4*b^2))/(a*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^
2)) + (2*tan(c/2 + (d*x)/2)*(3*a^6 - 3*b^6 + 22*a^2*b^4 + 13*a^4*b^2))/(3*a*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2
)) + (2*tan(c/2 + (d*x)/2)^5*(a^6 - 9*b^6 + 38*a^2*b^4 + 3*a^4*b^2))/(3*a*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2))
- (2*tan(c/2 + (d*x)/2)^3*(a^6 + 9*b^6 - 46*a^2*b^4 - 9*a^4*b^2))/(3*a*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)))/
(d*(a + 2*b*tan(c/2 + (d*x)/2) + 2*a*tan(c/2 + (d*x)/2)^2 - 2*a*tan(c/2 + (d*x)/2)^6 - a*tan(c/2 + (d*x)/2)^8
+ 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*x)/2)^7)) - (10*a*b^4*atanh((2*a^6*b
+ 2*b^7 + 6*a^2*b^5 + 6*a^4*b^3 - 2*a*tan(c/2 + (d*x)/2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))/(2*(a^2 + b^2)^(
7/2))))/(d*(a^2 + b^2)^(7/2))

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