3.182 \(\int \frac{\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=335 \[ \frac{2 (-1)^{2/3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac{2 a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}-b^{2/3}}}+\frac{2 \sqrt [3]{-1} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt{a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac{3 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac{3 x}{8 b} \]

[Out]

(3*x)/(8*b) + (2*(-1)^(2/3)*a^(5/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)
^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(7/3)*d) - (2*a^(5/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[
(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b^(7/3)*d) + (2*(-1)^(1/3)*a^(5/3)*ArcTan[(
(-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) + (-1)^(1/
3)*b^(2/3)]*b^(7/3)*d) + (a*Cos[c + d*x])/(b^2*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]*Sin[
c + d*x]^3)/(4*b*d)

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Rubi [A]  time = 0.699105, antiderivative size = 335, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3220, 2638, 2635, 8, 2660, 618, 204} \[ \frac{2 (-1)^{2/3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac{2 a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}-b^{2/3}}}+\frac{2 \sqrt [3]{-1} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt{a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt{a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac{3 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac{3 x}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a + b*Sin[c + d*x]^3),x]

[Out]

(3*x)/(8*b) + (2*(-1)^(2/3)*a^(5/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)
^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(7/3)*d) - (2*a^(5/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[
(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b^(7/3)*d) + (2*(-1)^(1/3)*a^(5/3)*ArcTan[(
(-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) + (-1)^(1/
3)*b^(2/3)]*b^(7/3)*d) + (a*Cos[c + d*x])/(b^2*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]*Sin[
c + d*x]^3)/(4*b*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [C]  time = 0.487818, size = 219, normalized size = 0.65 \[ \frac{-32 a^2 \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{-i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{\text{$\#$1}^4 b-2 \text{$\#$1}^2 b-4 i \text{$\#$1} a+b}\& \right ]+96 a \cos (c+d x)+3 b (12 (c+d x)-8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{96 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^7/(a + b*Sin[c + d*x]^3),x]

[Out]

(96*a*Cos[c + d*x] - 32*a^2*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (-2*ArcTan[
Sin[c + d*x]/(Cos[c + d*x] - #1)] + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x]
- #1)]*#1^2 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ] + 3*b*(12*(c
+ d*x) - 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(96*b^2*d)

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Maple [C]  time = 0.178, size = 366, normalized size = 1.1 \begin{align*}{\frac{2\,{a}^{2}}{3\,{b}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{3}+{\it \_R}}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }}+{\frac{3}{4\,bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{11}{4\,bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{11}{4\,bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{3}{4\,bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+2\,{\frac{a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{3}{4\,bd}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x)

[Out]

2/3/d*a^2/b^2*sum((_R^3+_R)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*
a+8*_Z^3*b+3*_Z^2*a+a))+3/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^
2)^4*tan(1/2*d*x+1/2*c)^6*a+11/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5+6/d/b^2/(1+tan(1/2*d*x+1/
2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*a-11/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+6/d/b^2/(1+tan(1/2*d
*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a-3/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+2/d/b^2/(1+tan(1/2
*d*x+1/2*c)^2)^4*a+3/4/d/b*arctan(tan(1/2*d*x+1/2*c))

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Maxima [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(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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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(sin(d*x+c)**7/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^7/(b*sin(d*x + c)^3 + a), x)