Integrand size = 23, antiderivative size = 73 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=-\frac {5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}}+\frac {5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right ) \sin ^{\frac {4}{5}}(c+d x)}{d \sqrt {\cos ^2(c+d x)}} \]
-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)
Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.51 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\frac {5 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (-9 \operatorname {Hypergeometric2F1}\left (-\frac {1}{10},\frac {1}{2},\frac {9}{10},\sin ^2(c+d x)\right )+\sin (c+d x) \left (9 \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right )+4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{10},\frac {19}{10},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{9 d \sqrt [5]{\sin (c+d x)}} \]
(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))
Time = 0.33 (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 |
\(\displaystyle \frac {5 \sin ^{\frac {4}{5}}(c+d x) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {1}{2},\frac {7}{5},\sin ^2(c+d x)\right )}{d \sqrt {\cos ^2(c+d x)}}-\frac {5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}}\) |
(-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])
3.3.20.3.1 Defintions of rubi rules used
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]
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]
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 ]
\[\int \frac {\left (2 \sin \left (d x +c \right )+1\right )^{2}}{\sin \left (d x +c \right )^{\frac {6}{5}}}d x\]
\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \]
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)
Timed out. \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \]
\[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=\int { \frac {{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac {6}{5}}} \,d x } \]
Time = 6.98 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.74 \[ \int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx=-\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{4/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{5};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{2/5}}-\frac {\cos \left (c+d\,x\right )\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{1/10}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{10};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\sin \left (c+d\,x\right )}^{1/5}}-\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{9/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{10},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{9/10}} \]
- (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))