3.6.5 \(\int \frac {\cos ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [505]

Optimal. Leaf size=261 \[ -\frac {b \left (a^2+4 b^2\right ) x}{a^5}+\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

-b*(a^2+4*b^2)*x/a^5+2*b^4*(5*a^2-4*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^5/(a-b)^(3/2)/(
a+b)^(3/2)/d+1/3*(2*a^4+7*a^2*b^2-12*b^4)*sin(d*x+c)/a^4/(a^2-b^2)/d-b*(a^2-2*b^2)*cos(d*x+c)*sin(d*x+c)/a^3/(
a^2-b^2)/d+1/3*(a^2-4*b^2)*cos(d*x+c)^2*sin(d*x+c)/a^2/(a^2-b^2)/d+b^2*cos(d*x+c)^2*sin(d*x+c)/a/(a^2-b^2)/d/(
a+b*sec(d*x+c))

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Rubi [A]
time = 0.58, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3932, 4189, 4004, 3916, 2738, 214} \begin {gather*} \frac {\left (a^2-4 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {b x \left (a^2+4 b^2\right )}{a^5}+\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac {b \left (a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(a^2 + 4*b^2)*x)/a^5) + (2*b^4*(5*a^2 - 4*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*
(a - b)^(3/2)*(a + b)^(3/2)*d) + ((2*a^4 + 7*a^2*b^2 - 12*b^4)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*d) - (b*(a^2 -
2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 4*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*(a
^2 - b^2)*d) + (b^2*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 214

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3932

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b^2*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)
*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4189

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\cos ^3(c+d x) \left (-a^2+4 b^2+a b \sec (c+d x)-3 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\cos ^2(c+d x) \left (-6 b \left (a^2-2 b^2\right )+a \left (2 a^2+b^2\right ) \sec (c+d x)+2 b \left (a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\cos (c+d x) \left (-2 \left (2 a^4+7 a^2 b^2-12 b^4\right )+2 a b \left (a^2+2 b^2\right ) \sec (c+d x)+6 b^2 \left (a^2-2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {-6 b \left (a^4+3 a^2 b^2-4 b^4\right )-6 a b^2 \left (a^2-2 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac {b \left (a^2+4 b^2\right ) x}{a^5}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (b^4 \left (5 a^2-4 b^2\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac {b \left (a^2+4 b^2\right ) x}{a^5}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (b^3 \left (5 a^2-4 b^2\right )\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac {b \left (a^2+4 b^2\right ) x}{a^5}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (2 b^3 \left (5 a^2-4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 \left (a^2-b^2\right ) d}\\ &=-\frac {b \left (a^2+4 b^2\right ) x}{a^5}+\frac {2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.14, size = 176, normalized size = 0.67 \begin {gather*} \frac {-12 b (-i a+2 b) (i a+2 b) (c+d x)+\frac {24 b^4 \left (-5 a^2+4 b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+9 a \left (a^2+4 b^2\right ) \sin (c+d x)+\frac {12 a b^5 \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))}-6 a^2 b \sin (2 (c+d x))+a^3 \sin (3 (c+d x))}{12 a^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*b*((-I)*a + 2*b)*(I*a + 2*b)*(c + d*x) + (24*b^4*(-5*a^2 + 4*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqr
t[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 9*a*(a^2 + 4*b^2)*Sin[c + d*x] + (12*a*b^5*Sin[c + d*x])/((-a + b)*(a + b)*
(b + a*Cos[c + d*x])) - 6*a^2*b*Sin[2*(c + d*x)] + a^3*Sin[3*(c + d*x)])/(12*a^5*d)

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Maple [A]
time = 0.21, size = 263, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (-a^{3}-b \,a^{2}-3 b^{2} a \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} a^{3}-6 b^{2} a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3}+b \,a^{2}-3 b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+b \left (a^{2}+4 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{5}}-\frac {2 b^{4} \left (-\frac {b a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (5 a^{2}-4 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5}}}{d}\)	\(263\)
default	\(\frac {-\frac {2 \left (\frac {\left (-a^{3}-b \,a^{2}-3 b^{2} a \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} a^{3}-6 b^{2} a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3}+b \,a^{2}-3 b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+b \left (a^{2}+4 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{5}}-\frac {2 b^{4} \left (-\frac {b a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (5 a^{2}-4 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5}}}{d}\)	\(263\)
risch	\(-\frac {b x}{a^{3}}-\frac {4 b^{3} x}{a^{5}}+\frac {i b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 a^{4} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 a^{4} d}-\frac {i b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{3} d}-\frac {2 i b^{5} \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{5} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {4 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{5}}-\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {4 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{5}}+\frac {\sin \left (3 d x +3 c \right )}{12 a^{2} d}\)	\(562\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [A]
time = 2.73, size = 757, normalized size = 2.90 \begin {gather*} \left [-\frac {6 \, {\left (a^{7} b + 2 \, a^{5} b^{3} - 7 \, a^{3} b^{5} + 4 \, a b^{7}\right )} d x \cos \left (d x + c\right ) + 6 \, {\left (a^{6} b^{2} + 2 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 4 \, b^{8}\right )} d x - 3 \, {\left (5 \, a^{2} b^{5} - 4 \, b^{7} + {\left (5 \, a^{3} b^{4} - 4 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (2 \, a^{7} b + 5 \, a^{5} b^{3} - 19 \, a^{3} b^{5} + 12 \, a b^{7} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} + a^{6} b^{2} - 5 \, a^{4} b^{4} + 3 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{10} - 2 \, a^{8} b^{2} + a^{6} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} b - 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d\right )}}, -\frac {3 \, {\left (a^{7} b + 2 \, a^{5} b^{3} - 7 \, a^{3} b^{5} + 4 \, a b^{7}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (a^{6} b^{2} + 2 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 4 \, b^{8}\right )} d x - 3 \, {\left (5 \, a^{2} b^{5} - 4 \, b^{7} + {\left (5 \, a^{3} b^{4} - 4 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{7} b + 5 \, a^{5} b^{3} - 19 \, a^{3} b^{5} + 12 \, a b^{7} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} + a^{6} b^{2} - 5 \, a^{4} b^{4} + 3 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{10} - 2 \, a^{8} b^{2} + a^{6} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} b - 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(6*(a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*d*x*cos(d*x + c) + 6*(a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b
^8)*d*x - 3*(5*a^2*b^5 - 4*b^7 + (5*a^3*b^4 - 4*a*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) -
(a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*
x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7 + (a^8 - 2*a^6*b^2 + a^
4*b^4)*cos(d*x + c)^3 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + 2*(a^8 + a^6*b^2 - 5*a^4*b^4 + 3*a^2*
b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b^3 + a^5*b^5)*
d), -1/3*(3*(a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*d*x*cos(d*x + c) + 3*(a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 +
4*b^8)*d*x - 3*(5*a^2*b^5 - 4*b^7 + (5*a^3*b^4 - 4*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b
^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7 + (a^8 - 2
*a^6*b^2 + a^4*b^4)*cos(d*x + c)^3 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + 2*(a^8 + a^6*b^2 - 5*a^4
*b^4 + 3*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b^
3 + a^5*b^5)*d)]

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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*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 335, normalized size = 1.28 \begin {gather*} \frac {\frac {6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - a^{4} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} + \frac {6 \, {\left (5 \, a^{2} b^{4} - 4 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{7} - a^{5} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {3 \, {\left (a^{2} b + 4 \, b^{3}\right )} {\left (d x + c\right )}}{a^{5}} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{4}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*b^5*tan(1/2*d*x + 1/2*c)/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)
) + 6*(5*a^2*b^4 - 4*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c)
- b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^7 - a^5*b^2)*sqrt(-a^2 + b^2)) - 3*(a^2*b + 4*b^3)*(d*x + c)/
a^5 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d*x + 1/2*c)^5 + 2*a^2*ta
n(1/2*d*x + 1/2*c)^3 + 18*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c)
+ 9*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^4))/d

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Mupad [B]
time = 9.08, size = 2500, normalized size = 9.58 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*tan(c/2 + (d*x)/2)^7*(a^5 - 2*a*b^4 + 4*b^5 - 3*a^2*b^3 + a^3*b^2))/(a^4*(a + b)*(a - b)) + (2*tan(c/2 +
(d*x)/2)^3*(6*a*b^4 - 8*a^4*b + a^5 + 36*b^5 - 19*a^2*b^3 - 7*a^3*b^2))/(3*a^4*(a + b)*(a - b)) - (2*tan(c/2
+ (d*x)/2)^5*(6*a*b^4 + 8*a^4*b + a^5 - 36*b^5 + 19*a^2*b^3 - 7*a^3*b^2))/(3*a^4*(a + b)*(a - b)) - (2*tan(c/2
+ (d*x)/2)*(a^5 - 2*a*b^4 - 4*b^5 + 3*a^2*b^3 + a^3*b^2))/(a^4*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/
2)^8*(a - b) + tan(c/2 + (d*x)/2)^2*(2*a + 4*b) - tan(c/2 + (d*x)/2)^6*(2*a - 4*b) + 6*b*tan(c/2 + (d*x)/2)^4)
) - (2*b*atan(((b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^
4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9
*b^2) + (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*
b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a^10*b^6 + 2*a
^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2)*32i)/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2)))*1i)/a^5))/a^5
+ (b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5
*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b*(a
^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b
+ a^15 - a^12*b^3 - a^13*b^2) + (b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a
^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2)*32i)/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2)))*1i)/a^5))/a^5)/((64*(64*b^1
4 - 32*a*b^13 - 112*a^2*b^12 + 48*a^3*b^11 + 12*a^4*b^10 - 6*a^5*b^9 + 31*a^6*b^8 - 5*a^7*b^7 + 5*a^8*b^6))/(a
^14*b + a^15 - a^12*b^3 - a^13*b^2) - (b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b
^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b
+ a^11 - a^8*b^3 - a^9*b^2) + (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b
^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^
15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2)*32i)/(a^5*(a^10*b + a^11 - a^8*b^3 - a^
9*b^2)))*1i)/a^5)*1i)/a^5 + (b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a
^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 -
a^8*b^3 - a^9*b^2) - (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^1
4*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a^15*b - 2*a
^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2)*32i)/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2)))*1
i)/a^5)*1i)/a^5))*(a^2 + 4*b^2))/(a^5*d) - (b^4*atan(((b^4*(5*a^2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*ta
n(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*
b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) + (b^4*((32*(a^17*b - 4*a^10*b^8
+ 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b
^4*tan(c/2 + (d*x)/2)*(5*a^2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12
*b^4 - 4*a^13*b^3 - 2*a^14*b^2))/((a^10*b + a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)
))*(5*a^2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))*1i)/(a^11 - a^5*b^6
+ 3*a^7*b^4 - 3*a^9*b^2) + (b^4*(5*a^2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 -
32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3
+ a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 -
4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) + (32*b^4*tan(c/2 + (d*x)/2)*(5*a^
2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*
b^2))/((a^10*b + a^11 - a^8*b^3 - a^9*b^2)*(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2)))*(5*a^2 - 4*b^2)*((a + b)
^3*(a - b)^3)^(1/2))/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))*1i)/(a^11 - a^5*b^6 + 3*a^7*b^4 - 3*a^9*b^2))/(
(64*(64*b^14 - 32*a*b^13 - 112*a^2*b^12 + 48*a^3*b^11 + 12*a^4*b^10 - 6*a^5*b^9 + 31*a^6*b^8 - 5*a^7*b^7 + 5*a
^8*b^6))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (b^4*(5*a^2 - 4*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2
+ (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 +
7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) + (b^4*((32*(a^17*b - 4*a^10*b^8 + 2*a
^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b^4*ta
n(c/2 + (d*x)/2)*(5*a^2 - 4*b^2)*((a + b)^3*(a ...

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