Integrand size = 23, antiderivative size = 194 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m)}+\frac {\left (b^2 (1+m)+a^2 (2+m)\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) (2+m) \sqrt {\cos ^2(e+f x)}}+\frac {2 a b \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) \sqrt {\cos ^2(e+f x)}} \]
-b^2*cos(f*x+e)*(d*sin(f*x+e))^(1+m)/d/f/(2+m)+(b^2*(1+m)+a^2*(2+m))*cos(f *x+e)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(f*x+e)^2)*(d*sin(f*x+e))^ (1+m)/d/f/(1+m)/(2+m)/(cos(f*x+e)^2)^(1/2)+2*a*b*cos(f*x+e)*hypergeom([1/2 , 1+1/2*m],[2+1/2*m],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+m)/d^2/f/(2+m)/(cos(f *x+e)^2)^(1/2)
Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.86 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\frac {\sqrt {\cos ^2(e+f x)} (d \sin (e+f x))^m \left (a^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right )+b (1+m) \sin (e+f x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right )+b (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right ) \tan (e+f x)}{f (1+m) (2+m) (3+m)} \]
(Sqrt[Cos[e + f*x]^2]*(d*Sin[e + f*x])^m*(a^2*(6 + 5*m + m^2)*Hypergeometr ic2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^2] + b*(1 + m)*Sin[e + f*x]* (2*a*(3 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2] + b*(2 + m)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Sin[e + f*x]^2]*S in[e + f*x]))*Tan[e + f*x])/(f*(1 + m)*(2 + m)*(3 + m))
Time = 0.54 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3268, 3042, 3122, 3493, 3042, 3122}
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.
\(\displaystyle \int (a+b \sin (e+f x))^2 (d \sin (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (e+f x))^2 (d \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3268 |
\(\displaystyle \int (d \sin (e+f x))^m \left (a^2+b^2 \sin ^2(e+f x)\right )dx+\frac {2 a b \int (d \sin (e+f x))^{m+1}dx}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sin (e+f x))^m \left (a^2+b^2 \sin (e+f x)^2\right )dx+\frac {2 a b \int (d \sin (e+f x))^{m+1}dx}{d}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \int (d \sin (e+f x))^m \left (a^2+b^2 \sin (e+f x)^2\right )dx+\frac {2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \int (d \sin (e+f x))^mdx+\frac {2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \int (d \sin (e+f x))^mdx+\frac {2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(e+f x)\right )}{d f (m+1) \sqrt {\cos ^2(e+f x)}}+\frac {2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)}\) |
-((b^2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + m))/(d*f*(2 + m))) + ((a^2 + (b^ 2*(1 + m))/(2 + m))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m) /2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(1 + m))/(d*f*(1 + m)*Sqrt[Cos[e + f* x]^2]) + (2*a*b*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + m))/(d^2*f*(2 + m)*Sqrt[Cos[e + f*x] ^2])
3.3.14.3.1 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[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{2}d x\]
\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\text {Timed out} \]
\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]