\(\int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \, dx\) [100]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {8\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{7 d (1+\sin (c+d x))^{5/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{7 d} \] Output:

-8/7*2^(5/6)*a*cos(d*x+c)*hypergeom([-5/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))* 
(a+a*sin(d*x+c))^(1/3)/d/(1+sin(d*x+c))^(5/6)-3/7*cos(d*x+c)*(a+a*sin(d*x+ 
c))^(4/3)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(97)=194\).

Time = 2.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.41 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\frac {(a (1+\sin (c+d x)))^{4/3} \left (40 \sqrt [3]{2} \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )-\sqrt {2-2 \sin (c+d x)} \left (20 \sqrt [3]{2} \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{2/3} (4 \cos (c+d x)+\sin (2 (c+d x))) \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}\right )\right )}{28 d \sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{8/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}} \] Input:

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

Output:

((a*(1 + Sin[c + d*x]))^(4/3)*(40*2^(1/3)*Cos[(2*c + Pi + 2*d*x)/4]*Hyperg 
eometricPFQ[{-1/2, -1/6}, {5/6}, Sin[(2*c + Pi + 2*d*x)/4]^2] - Sqrt[2 - 2 
*Sin[c + d*x]]*(20*2^(1/3)*Cos[(2*c + Pi + 2*d*x)/4] + 3*(Cos[(c + d*x)/2] 
 + Sin[(c + d*x)/2])^(2/3)*(4*Cos[c + d*x] + Sin[2*(c + d*x)])*Sin[(2*c + 
Pi + 2*d*x)/4]^(1/3))))/(28*d*Sqrt[Cos[(2*c + Pi + 2*d*x)/4]^2]*(Cos[(c + 
d*x)/2] + Sin[(c + d*x)/2])^(8/3)*Sin[(2*c + Pi + 2*d*x)/4]^(1/3))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {4}{7} \int (\sin (c+d x) a+a)^{4/3}dx-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{7} \int (\sin (c+d x) a+a)^{4/3}dx-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {4 a \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}\)

\(\Big \downarrow \) 3130

\(\displaystyle -\frac {8\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{7 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}\)

Input:

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

Output:

(-8*2^(5/6)*a*Cos[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 3/2, (1 - Sin[c + 
d*x])/2]*(a + a*Sin[c + d*x])^(1/3))/(7*d*(1 + Sin[c + d*x])^(5/6)) - (3*C 
os[c + d*x]*(a + a*Sin[c + d*x])^(4/3))/(7*d)
 

Defintions of rubi rules used

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)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a*(sin(c + d*x) + 1))**(4/3)*sin(c + d*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \, dx=a^{\frac {4}{3}} \left (\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \sin \left (d x +c \right )^{2}d x +\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \sin \left (d x +c \right )d x \right ) \] Input:

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

Output:

a**(1/3)*a*(int((sin(c + d*x) + 1)**(1/3)*sin(c + d*x)**2,x) + int((sin(c 
+ d*x) + 1)**(1/3)*sin(c + d*x),x))