\(\int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx\) [220]

Optimal result

Integrand size = 23, antiderivative size = 73 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=-\frac {5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}}+\frac {5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right ) \sin ^{\frac {4}{5}}(c+d x)}{d \sqrt {\cos ^2(c+d x)}} \] Output:

-5*cos(d*x+c)/d/sin(d*x+c)^(1/5)+5*cos(d*x+c)*hypergeom([2/5, 1/2],[7/5],s 
in(d*x+c)^2)*sin(d*x+c)^(4/5)/d/(cos(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.51 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\frac {5 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (-9 \operatorname {Hypergeometric2F1}\left (-\frac {1}{10},\frac {1}{2},\frac {9}{10},\sin ^2(c+d x)\right )+\sin (c+d x) \left (9 \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right )+4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{10},\frac {19}{10},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{9 d \sqrt [5]{\sin (c+d x)}} \] Input:

Integrate[(1 + 2*Sin[c + d*x])^2/Sin[c + d*x]^(6/5),x]
 

Output:

(5*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*(-9*Hypergeometric2F1[-1/10, 1/2, 9/1 
0, Sin[c + d*x]^2] + Sin[c + d*x]*(9*Hypergeometric2F1[2/5, 1/2, 7/5, Sin[ 
c + d*x]^2] + 4*Hypergeometric2F1[1/2, 9/10, 19/10, Sin[c + d*x]^2]*Sin[c 
+ d*x])))/(9*d*Sin[c + d*x]^(1/5))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3268, 3042, 3122, 3490}

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.

\(\displaystyle \int \frac {(2 \sin (c+d x)+1)^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(2 \sin (c+d x)+1)^2}{\sin (c+d x)^{6/5}}dx\)

\(\Big \downarrow \) 3268

\(\displaystyle \int \frac {4 \sin ^2(c+d x)+1}{\sin ^{\frac {6}{5}}(c+d x)}dx+4 \int \frac {1}{\sqrt [5]{\sin (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 \int \frac {1}{\sqrt [5]{\sin (c+d x)}}dx+\int \frac {4 \sin (c+d x)^2+1}{\sin (c+d x)^{6/5}}dx\)

\(\Big \downarrow \) 3122

\(\displaystyle \int \frac {4 \sin (c+d x)^2+1}{\sin (c+d x)^{6/5}}dx+\frac {5 \sin ^{\frac {4}{5}}(c+d x) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right )}{d \sqrt {\cos ^2(c+d x)}}\)

\(\Big \downarrow \) 3490

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

Input:

Int[(1 + 2*Sin[c + d*x])^2/Sin[c + d*x]^(6/5),x]
 

Output:

(-5*Cos[c + d*x])/(d*Sin[c + d*x]^(1/5)) + (5*Cos[c + d*x]*Hypergeometric2 
F1[2/5, 1/2, 7/5, Sin[c + d*x]^2]*Sin[c + d*x]^(4/5))/(d*Sqrt[Cos[c + d*x] 
^2])
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3268

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, x]
 

rule 3490

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

Maple [F]

\[\int \frac {\left (2 \sin \left (d x +c \right )+1\right )^{2}}{\sin \left (d x +c \right )^{\frac {6}{5}}}d x\]

Input:

int((2*sin(d*x+c)+1)^2/sin(d*x+c)^(6/5),x)
 

Output:

int((2*sin(d*x+c)+1)^2/sin(d*x+c)^(6/5),x)
 

Fricas [F]

\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \] Input:

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="fricas")
 

Output:

integral((4*cos(d*x + c)^2 - 4*sin(d*x + c) - 5)*sin(d*x + c)^(4/5)/(cos(d 
*x + c)^2 - 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((1+2*sin(d*x+c))**2/sin(d*x+c)**(6/5),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \] Input:

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="maxima")
 

Output:

integrate((2*sin(d*x + c) + 1)^2/sin(d*x + c)^(6/5), x)
 

Giac [F]

\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \] Input:

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="giac")
 

Output:

integrate((2*sin(d*x + c) + 1)^2/sin(d*x + c)^(6/5), x)
 

Mupad [B] (verification not implemented)

Time = 18.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.74 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=-\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{4/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{5};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{2/5}}-\frac {\cos \left (c+d\,x\right )\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{1/10}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{10};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\sin \left (c+d\,x\right )}^{1/5}}-\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{9/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{10},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{9/10}} \] Input:

int((2*sin(c + d*x) + 1)^2/sin(c + d*x)^(6/5),x)
 

Output:

- (4*cos(c + d*x)*sin(c + d*x)^(4/5)*hypergeom([1/2, 3/5], 3/2, cos(c + d* 
x)^2))/(d*(sin(c + d*x)^2)^(2/5)) - (cos(c + d*x)*(sin(c + d*x)^2)^(1/10)* 
hypergeom([1/2, 11/10], 3/2, cos(c + d*x)^2))/(d*sin(c + d*x)^(1/5)) - (4* 
cos(c + d*x)*sin(c + d*x)^(9/5)*hypergeom([1/10, 1/2], 3/2, cos(c + d*x)^2 
))/(d*(sin(c + d*x)^2)^(9/10))
 

Reduce [F]

\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=4 \left (\int \sin \left (d x +c \right )^{\frac {4}{5}}d x \right )+4 \left (\int \frac {1}{\sin \left (d x +c \right )^{\frac {1}{5}}}d x \right )+\int \frac {1}{\sin \left (d x +c \right )^{\frac {6}{5}}}d x \] Input:

int((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x)
 

Output:

4*int(sin(c + d*x)/sin(c + d*x)**(1/5),x) + 4*int(1/sin(c + d*x)**(1/5),x) 
 + int(1/(sin(c + d*x)**(1/5)*sin(c + d*x)),x)