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

Optimal. Leaf size=202 \[ \frac {a b \left (a^2-11 b^2\right )}{2 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac {b \left (a^2-2 b^2\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {a x \left (a^4+10 a^2 b^2-15 b^4\right )}{2 \left (a^2+b^2\right )^4} \]

[Out]

1/2*a*(a^4+10*a^2*b^2-15*b^4)*x/(a^2+b^2)^4+2*b^3*(5*a^2-b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^4/d+1/2*
b*(a^2-2*b^2)/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^2+1/2*cos(d*x+c)^2*(b+a*tan(d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x+c))^
2+1/2*a*b*(a^2-11*b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3506, 741, 801, 635, 203, 260} \[ \frac {a b \left (a^2-11 b^2\right )}{2 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac {b \left (a^2-2 b^2\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {a x \left (10 a^2 b^2+a^4-15 b^4\right )}{2 \left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^4 + 10*a^2*b^2 - 15*b^4)*x)/(2*(a^2 + b^2)^4) + (2*b^3*(5*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]
])/((a^2 + b^2)^4*d) + (b*(a^2 - 2*b^2))/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])^2) + (Cos[c + d*x]^2*(b + a*T
an[c + d*x]))/(2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*b*(a^2 - 11*b^2))/(2*(a^2 + b^2)^3*d*(a + b*Tan[c
+ d*x]))

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

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

Rule 741

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 3506

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{(a+x)^3 \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {b \operatorname {Subst}\left (\int \frac {-4-\frac {a^2}{b^2}-\frac {3 a x}{b^2}}{(a+x)^3 \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {b \operatorname {Subst}\left (\int \left (\frac {2 \left (a^2-2 b^2\right )}{\left (a^2+b^2\right ) (a+x)^3}+\frac {a^3-11 a b^2}{\left (a^2+b^2\right )^2 (a+x)^2}+\frac {4 b^2 \left (-5 a^2+b^2\right )}{\left (a^2+b^2\right )^3 (a+x)}+\frac {-a \left (a^4+10 a^2 b^2-15 b^4\right )+4 b^2 \left (5 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {b \left (a^2-2 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a b \left (a^2-11 b^2\right )}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac {b \operatorname {Subst}\left (\int \frac {-a \left (a^4+10 a^2 b^2-15 b^4\right )+4 b^2 \left (5 a^2-b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 d}\\ &=\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {b \left (a^2-2 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a b \left (a^2-11 b^2\right )}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac {\left (2 b^3 \left (5 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^4 d}+\frac {\left (a b \left (a^4+10 a^2 b^2-15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 d}\\ &=\frac {a \left (a^4+10 a^2 b^2-15 b^4\right ) x}{2 \left (a^2+b^2\right )^4}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {b \left (a^2-2 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a b \left (a^2-11 b^2\right )}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 6.28, size = 458, normalized size = 2.27 \[ \frac {b^3 \left (\frac {\cos ^2(c+d x) \left (a b \tan (c+d x)+b^2\right )}{2 b^4 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\left (2 a^2-4 b^2\right ) \left (-\frac {2 a}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {1}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}-\frac {\left (-\frac {a^3-3 a b^2}{\sqrt {-b^2}}+3 a^2-b^2\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3}-\frac {\left (\frac {a^3-3 a b^2}{\sqrt {-b^2}}+3 a^2-b^2\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3}\right )-3 a \left (-\frac {1}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (2 a-\frac {a^2-b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2}+\frac {2 a \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}-\frac {\left (\frac {a^2-b^2}{\sqrt {-b^2}}+2 a\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2}\right )}{2 b^2 \left (a^2+b^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^3*((Cos[c + d*x]^2*(b^2 + a*b*Tan[c + d*x]))/(2*b^4*(a^2 + b^2)*(a + b*Tan[c + d*x])^2) - ((2*a^2 - 4*b^2)*
(-1/2*((3*a^2 - b^2 - (a^3 - 3*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/(a^2 + b^2)^3 + ((3*a^2 -
b^2)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^3 - ((3*a^2 - b^2 + (a^3 - 3*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*T
an[c + d*x]])/(2*(a^2 + b^2)^3) - 1/(2*(a^2 + b^2)*(a + b*Tan[c + d*x])^2) - (2*a)/((a^2 + b^2)^2*(a + b*Tan[c
+ d*x]))) - 3*a*(-1/2*((2*a - (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/(a^2 + b^2)^2 + (2*a*
Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^2 - ((2*a + (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/(2*
(a^2 + b^2)^2) - 1/((a^2 + b^2)*(a + b*Tan[c + d*x]))))/(2*b^2*(a^2 + b^2))))/d

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fricas [B]  time = 1.12, size = 503, normalized size = 2.49 \[ -\frac {3 \, a^{4} b^{3} - 16 \, a^{2} b^{5} + b^{7} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b^{2} + 10 \, a^{3} b^{4} - 15 \, a b^{6}\right )} d x - {\left (a^{6} b - a^{4} b^{3} - 45 \, a^{2} b^{5} - 3 \, b^{7} + 2 \, {\left (a^{7} + 9 \, a^{5} b^{2} - 25 \, a^{3} b^{4} + 15 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, a^{2} b^{5} - b^{7} + {\left (5 \, a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b^{2} - 3 \, a^{3} b^{4} + 6 \, a b^{6} - {\left (a^{6} b + 10 \, a^{4} b^{3} - 15 \, a^{2} b^{5}\right )} d x\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*a^4*b^3 - 16*a^2*b^5 + 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^5*b^2 + 10*
a^3*b^4 - 15*a*b^6)*d*x - (a^6*b - a^4*b^3 - 45*a^2*b^5 - 3*b^7 + 2*(a^7 + 9*a^5*b^2 - 25*a^3*b^4 + 15*a*b^6)*
d*x)*cos(d*x + c)^2 - 4*(5*a^2*b^5 - b^7 + (5*a^4*b^3 - 6*a^2*b^5 + b^7)*cos(d*x + c)^2 + 2*(5*a^3*b^4 - a*b^6
)*cos(d*x + c)*sin(d*x + c))*log(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^5*b^2 - 3*a^3*b^4 + 6*a*b^6 - (a^6*b + 10*a^4*b^3 - 15
*a^2*b^5)*d*x)*cos(d*x + c))*sin(d*x + c))/((a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)*d*co
s(d*x + c)^2 + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^8*b^2 +
4*a^6*b^4 + 6*a^4*b^6 + 4*a^2*b^8 + b^10)*d)

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giac [B]  time = 1.14, size = 439, normalized size = 2.17 \[ \frac {\frac {{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {2 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {4 \, {\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} + \frac {10 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 2 \, b^{5} \tan \left (d x + c\right )^{2} + a^{5} \tan \left (d x + c\right ) - 2 \, a^{3} b^{2} \tan \left (d x + c\right ) - 3 \, a b^{4} \tan \left (d x + c\right ) + 3 \, a^{4} b + 12 \, a^{2} b^{3} - 3 \, b^{5}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}} - \frac {30 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} - 6 \, b^{7} \tan \left (d x + c\right )^{2} + 68 \, a^{3} b^{4} \tan \left (d x + c\right ) - 4 \, a b^{6} \tan \left (d x + c\right ) + 39 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^5 + 10*a^3*b^2 - 15*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 2*(5*a^2*b^3 -
b^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 4*(5*a^2*b^4 - b^6)*log(abs(b*t
an(d*x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) + (10*a^2*b^3*tan(d*x + c)^2 - 2*b^5*tan(d
*x + c)^2 + a^5*tan(d*x + c) - 2*a^3*b^2*tan(d*x + c) - 3*a*b^4*tan(d*x + c) + 3*a^4*b + 12*a^2*b^3 - 3*b^5)/(
(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(d*x + c)^2 + 1)) - (30*a^2*b^5*tan(d*x + c)^2 - 6*b^7*tan
(d*x + c)^2 + 68*a^3*b^4*tan(d*x + c) - 4*a*b^6*tan(d*x + c) + 39*a^4*b^3 + 4*a^2*b^5 + b^7)/((a^8 + 4*a^6*b^2
+ 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^2))/d

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maple [B]  time = 0.52, size = 453, normalized size = 2.24 \[ -\frac {b^{3}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {4 b^{3} a}{d \left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {10 b^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{4}}-\frac {2 b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {\tan \left (d x +c \right ) a^{5}}{2 d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {\tan \left (d x +c \right ) b^{2} a^{3}}{d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {3 \tan \left (d x +c \right ) a \,b^{4}}{2 d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {3 a^{4} b}{2 d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {a^{2} b^{3}}{d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {b^{5}}{2 d \left (a^{2}+b^{2}\right )^{4} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{5}}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right ) b^{2} a^{3}}{d \left (a^{2}+b^{2}\right )^{4}}-\frac {15 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{4}}{2 d \left (a^{2}+b^{2}\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{5}}{2 d \left (a^{2}+b^{2}\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*b^3/(a^2+b^2)^2/(a+b*tan(d*x+c))^2-4/d*b^3/(a^2+b^2)^3*a/(a+b*tan(d*x+c))+10/d*b^3/(a^2+b^2)^4*ln(a+b*t
an(d*x+c))*a^2-2/d*b^5/(a^2+b^2)^4*ln(a+b*tan(d*x+c))+1/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*a^5-1/d/(a
^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*b^2*a^3-3/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*a*b^4+3/2/d/(a^2+b
^2)^4/(1+tan(d*x+c)^2)*a^4*b+1/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*a^2*b^3-1/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*b^5-5
/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*a^2*b^3+1/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*b^5+5/d/(a^2+b^2)^4*arctan(tan(d*
x+c))*b^2*a^3-15/2/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a*b^4+1/2/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^5

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maxima [B]  time = 0.46, size = 458, normalized size = 2.27 \[ \frac {\frac {{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {4 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {2 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, a^{4} b - 10 \, a^{2} b^{3} - b^{5} + {\left (a^{3} b^{2} - 11 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{5} + 3 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \tan \left (d x + c\right )}{a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^5 + 10*a^3*b^2 - 15*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 4*(5*a^2*b^3 -
b^5)*log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 2*(5*a^2*b^3 - b^5)*log(tan(d*x
+ c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + (3*a^4*b - 10*a^2*b^3 - b^5 + (a^3*b^2 - 11*a*b
^4)*tan(d*x + c)^3 + 2*(a^4*b - 6*a^2*b^3 - b^5)*tan(d*x + c)^2 + (a^5 + 3*a^3*b^2 - 10*a*b^4)*tan(d*x + c))/(
a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*tan(d*x + c)^4 + 2*(a^7*b + 3*
a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)^3 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*tan(d*x + c)^2 +
2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)))/d

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mupad [B]  time = 4.56, size = 419, normalized size = 2.07 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {10\,b^3}{{\left (a^2+b^2\right )}^3}-\frac {12\,b^5}{{\left (a^2+b^2\right )}^4}\right )}{d}-\frac {\frac {-3\,a^4\,b+10\,a^2\,b^3+b^5}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (11\,a\,b^4-a^3\,b^2\right )}{2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-a^4\,b+6\,a^2\,b^3+b^5\right )}{\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+3\,a^2\,b^2-10\,b^4\right )}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+b^2\right )+a^2+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (b+\frac {a\,1{}\mathrm {i}}{4}\right )}{d\,\left (a^4-a^3\,b\,4{}\mathrm {i}-6\,a^2\,b^2+a\,b^3\,4{}\mathrm {i}+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a+b\,4{}\mathrm {i}\right )}{4\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a + b*tan(c + d*x))*((10*b^3)/(a^2 + b^2)^3 - (12*b^5)/(a^2 + b^2)^4))/d - ((b^5 - 3*a^4*b + 10*a^2*b^3)/
(2*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)^3*(11*a*b^4 - a^3*b^2))/(2*(a^6 + b^6 + 3*a^2*b^4 + 3*
a^4*b^2)) + (tan(c + d*x)^2*(b^5 - a^4*b + 6*a^2*b^3))/((a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) - (a*tan(c + d*x)
*(a^4 - 10*b^4 + 3*a^2*b^2))/(2*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)))/(d*(tan(c + d*x)^2*(a^2 + b^2) + a^2 + b
^2*tan(c + d*x)^4 + 2*a*b*tan(c + d*x) + 2*a*b*tan(c + d*x)^3)) + (log(tan(c + d*x) + 1i)*((a*1i)/4 + b))/(d*(
a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)) + (log(tan(c + d*x) - 1i)*(a + b*4i))/(4*d*(4*a*b^3 - 4*a^3*b +
a^4*1i + b^4*1i - a^2*b^2*6i))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: AttributeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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