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

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

[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)

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Rubi [A]  time = 0.12543, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3195, 302, 205} \[ \frac{a^2 \tan ^3(c+d x)}{3 d (a+b)^3}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{9/2}}-\frac{a^3 \tan (c+d x)}{d (a+b)^4}+\frac{\tan ^7(c+d x)}{7 d (a+b)}-\frac{a \tan ^5(c+d x)}{5 d (a+b)^2} \]

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 3195

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]

Rule 302

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 205

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

Rubi steps

\begin{align*} \int \frac{\tan ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{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 \operatorname{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*}

Mathematica [A]  time = 2.40066, size = 147, normalized size = 1.22 \[ \frac{\tan (c+d x) \left (\left (254 a^2 b+122 a^3+177 a b^2+45 b^3\right ) \sec ^2(c+d x)-122 a^2 b-176 a^3-66 a b^2+15 (a+b)^3 \sec ^6(c+d x)-3 (a+b)^2 (22 a+15 b) \sec ^4(c+d x)-15 b^3\right )}{105 d (a+b)^4}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{9/2}} \]

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 [B]  time = 0.118, size = 252, normalized size = 2.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{7}{a}^{3}}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{3\, \left ( \tan \left ( dx+c \right ) \right ) ^{7}{a}^{2}b}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( a+b \right ) ^{4}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{3}}{5\,d \left ( a+b \right ) ^{4}}}-{\frac{2\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{2}b}{5\,d \left ( a+b \right ) ^{4}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}a{b}^{2}}{5\,d \left ( a+b \right ) ^{4}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{3}}{3\,d \left ( a+b \right ) ^{4}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{2}b}{3\,d \left ( a+b \right ) ^{4}}}-{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d \left ( a+b \right ) ^{4}}}+{\frac{{a}^{4}}{d \left ( a+b \right ) ^{4}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

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)

[Out]

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

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

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]

Exception raised: ValueError

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Fricas [B]  time = 2.28648, size = 1453, normalized size = 12.11 \begin{align*} \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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 9.94162, size = 637, normalized size = 5.31 \begin{align*} \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{align*}

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