3.5.49 \(\int \frac {\tan ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [449]

Optimal. Leaf size=120 \[ \frac {a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{9/2} d}-\frac {a^3 \tan (c+d x)}{(a+b)^4 d}+\frac {a^2 \tan ^3(c+d x)}{3 (a+b)^3 d}-\frac {a \tan ^5(c+d x)}{5 (a+b)^2 d}+\frac {\tan ^7(c+d x)}{7 (a+b) d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3274, 308, 211} \begin {gather*} \frac {a^{7/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{d (a+b)^{9/2}}-\frac {a^3 \tan (c+d x)}{d (a+b)^4}+\frac {a^2 \tan ^3(c+d x)}{3 d (a+b)^3}+\frac {\tan ^7(c+d x)}{7 d (a+b)}-\frac {a \tan ^5(c+d x)}{5 d (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

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

Rule 211

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 3274

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(p
+ 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 1.59, size = 147, normalized size = 1.22 \begin {gather*} \frac {a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{9/2} d}+\frac {\left (-176 a^3-122 a^2 b-66 a b^2-15 b^3+\left (122 a^3+254 a^2 b+177 a b^2+45 b^3\right ) \sec ^2(c+d x)-3 (a+b)^2 (22 a+15 b) \sec ^4(c+d x)+15 (a+b)^3 \sec ^6(c+d x)\right ) \tan (c+d x)}{105 (a+b)^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

(a^(7/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/((a + b)^(9/2)*d) + ((-176*a^3 - 122*a^2*b - 66*a*b^2 - 1
5*b^3 + (122*a^3 + 254*a^2*b + 177*a*b^2 + 45*b^3)*Sec[c + d*x]^2 - 3*(a + b)^2*(22*a + 15*b)*Sec[c + d*x]^4 +
15*(a + b)^3*Sec[c + d*x]^6)*Tan[c + d*x])/(105*(a + b)^4*d)

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Maple [A]
time = 0.66, size = 120, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {a \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right ) \left (a +b \right )}{3}-a^{3} \tan \left (d x +c \right )}{\left (a +b \right )^{4}}+\frac {a^{4} \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{4} \sqrt {a \left (a +b \right )}}}{d}\)	\(120\)
default	\(\frac {\frac {\frac {\left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {a \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right ) \left (a +b \right )}{3}-a^{3} \tan \left (d x +c \right )}{\left (a +b \right )^{4}}+\frac {a^{4} \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{4} \sqrt {a \left (a +b \right )}}}{d}\)	\(120\)
risch	\(-\frac {2 i \left (224 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}+66 a \,b^{2}+15 b^{3}+176 a^{3}+122 a^{2} b +315 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+420 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+1176 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+1722 b \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2}+42 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+1400 b \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}+3080 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+2436 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+812 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+1260 a^{3} {\mathrm e}^{10 i \left (d x +c \right )}+420 a^{3} {\mathrm e}^{12 i \left (d x +c \right )}+630 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}+420 a \,b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+840 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}+210 a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+2870 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+1890 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}+3080 a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+525 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}\right )}{105 d \left (a +b \right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}+\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{5} d}-\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{5} d}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^8/(a+sin(d*x+c)^2*b),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/(a+b)^4*(1/7*(a+b)*(a^2+2*a*b+b^2)*tan(d*x+c)^7-1/5*a*(a^2+2*a*b+b^2)*tan(d*x+c)^5+1/3*a^2*tan(d*x+c)^3
*(a+b)-a^3*tan(d*x+c))+a^4/(a+b)^4/(a*(a+b))^(1/2)*arctan(tan(d*x+c)*(a+b)/(a*(a+b))^(1/2)))

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Maxima [A]
time = 0.54, size = 180, normalized size = 1.50 \begin {gather*} \frac {\frac {105 \, a^{4} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{7} - 21 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{5} - 105 \, a^{3} \tan \left (d x + c\right ) + 35 \, {\left (a^{3} + a^{2} b\right )} \tan \left (d x + c\right )^{3}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/105*(105*a^4*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(
(a + b)*a)) + (15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(d*x + c)^7 - 21*(a^3 + 2*a^2*b + a*b^2)*tan(d*x + c)^5 -
105*a^3*tan(d*x + c) + 35*(a^3 + a^2*b)*tan(d*x + c)^3)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (106) = 212\).
time = 0.45, size = 602, normalized size = 5.02 \begin {gather*} \left [\frac {105 \, a^{3} \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{7} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left ({\left (176 \, a^{3} + 122 \, a^{2} b + 66 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - {\left (122 \, a^{3} + 254 \, a^{2} b + 177 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 15 \, a^{3} - 45 \, a^{2} b - 45 \, a b^{2} - 15 \, b^{3} + 3 \, {\left (22 \, a^{3} + 59 \, a^{2} b + 52 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{7}}, -\frac {105 \, a^{3} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{7} + 2 \, {\left ({\left (176 \, a^{3} + 122 \, a^{2} b + 66 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - {\left (122 \, a^{3} + 254 \, a^{2} b + 177 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 15 \, a^{3} - 45 \, a^{2} b - 45 \, a b^{2} - 15 \, b^{3} + 3 \, {\left (22 \, a^{3} + 59 \, a^{2} b + 52 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{7}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/420*(105*a^3*sqrt(-a/(a + b))*cos(d*x + c)^7*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b +
b^2)*cos(d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a
+ b))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2
)) - 4*((176*a^3 + 122*a^2*b + 66*a*b^2 + 15*b^3)*cos(d*x + c)^6 - (122*a^3 + 254*a^2*b + 177*a*b^2 + 45*b^3)*
cos(d*x + c)^4 - 15*a^3 - 45*a^2*b - 45*a*b^2 - 15*b^3 + 3*(22*a^3 + 59*a^2*b + 52*a*b^2 + 15*b^3)*cos(d*x + c
)^2)*sin(d*x + c))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*cos(d*x + c)^7), -1/210*(105*a^3*sqrt(a/(a +
b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c)
^7 + 2*((176*a^3 + 122*a^2*b + 66*a*b^2 + 15*b^3)*cos(d*x + c)^6 - (122*a^3 + 254*a^2*b + 177*a*b^2 + 45*b^3)*
cos(d*x + c)^4 - 15*a^3 - 45*a^2*b - 45*a*b^2 - 15*b^3 + 3*(22*a^3 + 59*a^2*b + 52*a*b^2 + 15*b^3)*cos(d*x + c
)^2)*sin(d*x + c))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*cos(d*x + c)^7)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{8}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**8/(a+b*sin(d*x+c)**2),x)

[Out]

Integral(tan(c + d*x)**8/(a + b*sin(c + d*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (106) = 212\).
time = 4.93, size = 472, normalized size = 3.93 \begin {gather*} \frac {\frac {105 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{4}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a^{2} + a b}} + \frac {15 \, a^{6} \tan \left (d x + c\right )^{7} + 90 \, a^{5} b \tan \left (d x + c\right )^{7} + 225 \, a^{4} b^{2} \tan \left (d x + c\right )^{7} + 300 \, a^{3} b^{3} \tan \left (d x + c\right )^{7} + 225 \, a^{2} b^{4} \tan \left (d x + c\right )^{7} + 90 \, a b^{5} \tan \left (d x + c\right )^{7} + 15 \, b^{6} \tan \left (d x + c\right )^{7} - 21 \, a^{6} \tan \left (d x + c\right )^{5} - 105 \, a^{5} b \tan \left (d x + c\right )^{5} - 210 \, a^{4} b^{2} \tan \left (d x + c\right )^{5} - 210 \, a^{3} b^{3} \tan \left (d x + c\right )^{5} - 105 \, a^{2} b^{4} \tan \left (d x + c\right )^{5} - 21 \, a b^{5} \tan \left (d x + c\right )^{5} + 35 \, a^{6} \tan \left (d x + c\right )^{3} + 140 \, a^{5} b \tan \left (d x + c\right )^{3} + 210 \, a^{4} b^{2} \tan \left (d x + c\right )^{3} + 140 \, a^{3} b^{3} \tan \left (d x + c\right )^{3} + 35 \, a^{2} b^{4} \tan \left (d x + c\right )^{3} - 105 \, a^{6} \tan \left (d x + c\right ) - 315 \, a^{5} b \tan \left (d x + c\right ) - 315 \, a^{4} b^{2} \tan \left (d x + c\right ) - 105 \, a^{3} b^{3} \tan \left (d x + c\right )}{a^{7} + 7 \, a^{6} b + 21 \, a^{5} b^{2} + 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 21 \, a^{2} b^{5} + 7 \, a b^{6} + b^{7}}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/105*(105*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 +
a*b)))*a^4/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a^2 + a*b)) + (15*a^6*tan(d*x + c)^7 + 90*a^5*b*t
an(d*x + c)^7 + 225*a^4*b^2*tan(d*x + c)^7 + 300*a^3*b^3*tan(d*x + c)^7 + 225*a^2*b^4*tan(d*x + c)^7 + 90*a*b^
5*tan(d*x + c)^7 + 15*b^6*tan(d*x + c)^7 - 21*a^6*tan(d*x + c)^5 - 105*a^5*b*tan(d*x + c)^5 - 210*a^4*b^2*tan(
d*x + c)^5 - 210*a^3*b^3*tan(d*x + c)^5 - 105*a^2*b^4*tan(d*x + c)^5 - 21*a*b^5*tan(d*x + c)^5 + 35*a^6*tan(d*
x + c)^3 + 140*a^5*b*tan(d*x + c)^3 + 210*a^4*b^2*tan(d*x + c)^3 + 140*a^3*b^3*tan(d*x + c)^3 + 35*a^2*b^4*tan
(d*x + c)^3 - 105*a^6*tan(d*x + c) - 315*a^5*b*tan(d*x + c) - 315*a^4*b^2*tan(d*x + c) - 105*a^3*b^3*tan(d*x +
c))/(a^7 + 7*a^6*b + 21*a^5*b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2*b^5 + 7*a*b^6 + b^7))/d

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Mupad [B]
time = 14.78, size = 141, normalized size = 1.18 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^7}{7\,d\,\left (a+b\right )}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,{\left (a+b\right )}^3}+\frac {a^{7/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{9/2}}\right )}{d\,{\left (a+b\right )}^{9/2}}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d\,{\left (a+b\right )}^2}-\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^8/(a + b*sin(c + d*x)^2),x)

[Out]

tan(c + d*x)^7/(7*d*(a + b)) + (a^2*tan(c + d*x)^3)/(3*d*(a + b)^3) + (a^(7/2)*atan((tan(c + d*x)*(2*a + 2*b)*
(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2))/(2*a^(1/2)*(a + b)^(9/2))))/(d*(a + b)^(9/2)) - (a*tan(c + d*x)^5
)/(5*d*(a + b)^2) - (a^3*tan(c + d*x))/(d*(a + b)^4)

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