\(\int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx\) [110]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a+a \sin (c+d x)}}-\frac {2 \sqrt [6]{2} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}} \] Output:

6/5*cos(d*x+c)/d/(a+a*sin(d*x+c))^(4/3)-3/5*cos(d*x+c)*sin(d*x+c)^2/d/(a+a 
*sin(d*x+c))^(4/3)+6/5*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/3)-2*2^(1/6)*cos 
(d*x+c)*hypergeom([1/2, 5/6],[3/2],1/2-1/2*sin(d*x+c))/a/d/(1+sin(d*x+c))^ 
(1/6)/(a+a*sin(d*x+c))^(1/3)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {3 \cos (c+d x) \left (20 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right ) (1+\sin (c+d x))+\sqrt {1-\sin (c+d x)} (7+\cos (2 (c+d x))+4 \sin (c+d x))\right )}{10 d \sqrt {1-\sin (c+d x)} (a (1+\sin (c+d x)))^{4/3}} \] Input:

Integrate[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(4/3),x]
 

Output:

(3*Cos[c + d*x]*(20*Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi 
 + 2*d*x)/4]^2]*(1 + Sin[c + d*x]) + Sqrt[1 - Sin[c + d*x]]*(7 + Cos[2*(c 
+ d*x)] + 4*Sin[c + d*x])))/(10*d*Sqrt[1 - Sin[c + d*x]]*(a*(1 + Sin[c + d 
*x]))^(4/3))
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3262, 27, 3042, 3447, 3042, 3498, 27, 3042, 3230, 3042, 3131, 3042, 3130}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^3(c+d x)}{(a \sin (c+d x)+a)^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3}{(a \sin (c+d x)+a)^{4/3}}dx\)

\(\Big \downarrow \) 3262

\(\displaystyle \frac {3 \int \frac {2 \sin (c+d x) (3 a-2 a \sin (c+d x))}{3 (\sin (c+d x) a+a)^{4/3}}dx}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int \frac {\sin (c+d x) (3 a-2 a \sin (c+d x))}{(\sin (c+d x) a+a)^{4/3}}dx}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {\sin (c+d x) (3 a-2 a \sin (c+d x))}{(\sin (c+d x) a+a)^{4/3}}dx}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {2 \int \frac {3 a \sin (c+d x)-2 a \sin ^2(c+d x)}{(\sin (c+d x) a+a)^{4/3}}dx}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {3 a \sin (c+d x)-2 a \sin (c+d x)^2}{(\sin (c+d x) a+a)^{4/3}}dx}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3498

\(\displaystyle \frac {2 \left (\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}-\frac {3 \int -\frac {10 \left (2 a^2-a^2 \sin (c+d x)\right )}{3 \sqrt [3]{\sin (c+d x) a+a}}dx}{5 a^2}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {2 a^2-a^2 \sin (c+d x)}{\sqrt [3]{\sin (c+d x) a+a}}dx}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {2 a^2-a^2 \sin (c+d x)}{\sqrt [3]{\sin (c+d x) a+a}}dx}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {5}{2} a^2 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx+\frac {3 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\right )}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {5}{2} a^2 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx+\frac {3 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\right )}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {5 a^2 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}+\frac {3 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\right )}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {5 a^2 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}+\frac {3 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\right )}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {3 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}-\frac {5 a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}\right )}{a^2}+\frac {3 a \cos (c+d x)}{d (a \sin (c+d x)+a)^{4/3}}\right )}{5 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\)

Input:

Int[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(4/3),x]
 

Output:

(-3*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*(a + a*Sin[c + d*x])^(4/3)) + (2*((3 
*a*Cos[c + d*x])/(d*(a + a*Sin[c + d*x])^(4/3)) + (2*((3*a^2*Cos[c + d*x]) 
/(2*d*(a + a*Sin[c + d*x])^(1/3)) - (5*a^2*Cos[c + d*x]*Hypergeometric2F1[ 
1/2, 5/6, 3/2, (1 - Sin[c + d*x])/2])/(2^(5/6)*d*(1 + Sin[c + d*x])^(1/6)* 
(a + a*Sin[c + d*x])^(1/3))))/a^2))/(5*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3230

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 3262

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x]) 
^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[1/(b*(m + n))   In 
t[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 2)*Simp[d*(a*c*m + b*d*( 
n - 1)) + b*c^2*(m + n) + d*(a*d*m + b*c*(m + 2*n - 1))*Sin[e + f*x], x], x 
], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3498

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* 
B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 
/(a^2*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b 
*B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
 B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \frac {\sin \left (d x +c \right )^{3}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

Input:

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x)
 

Output:

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x)
 

Fricas [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

integral((cos(d*x + c)^2 - 1)*(a*sin(d*x + c) + a)^(2/3)*sin(d*x + c)/(a^2 
*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**3/(a+a*sin(d*x+c))**(4/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(4/3), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:

int(sin(c + d*x)^3/(a + a*sin(c + d*x))^(4/3),x)
 

Output:

int(sin(c + d*x)^3/(a + a*sin(c + d*x))^(4/3), x)
 

Reduce [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {\int \frac {\sin \left (d x +c \right )^{3}}{\left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}}}d x}{a^{\frac {4}{3}}} \] Input:

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x)
 

Output:

int(sin(c + d*x)**3/((sin(c + d*x) + 1)**(1/3)*sin(c + d*x) + (sin(c + d*x 
) + 1)**(1/3)),x)/(a**(1/3)*a)