\(\int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx\) [214]

Optimal result

Integrand size = 23, antiderivative size = 194 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m)}+\frac {\left (b^2 (1+m)+a^2 (2+m)\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) (2+m) \sqrt {\cos ^2(e+f x)}}+\frac {2 a b \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) \sqrt {\cos ^2(e+f x)}} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2868, 2722, 3093} \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\frac {\left (a^2 (m+2)+b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt {\cos ^2(e+f x)}}+\frac {2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)} \]

[In]

Int[(d*Sin[e + f*x])^m*(a + b*Sin[e + f*x])^2,x]

[Out]

-((b^2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + m))/(d*f*(2 + m))) + ((b^2*(1 + m) + a^2*(2 + m))*Cos[e + f*x]*Hyper
geometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(1 + m))/(d*f*(1 + m)*(2 + m)*Sqrt[Co
s[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])

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2868

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[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 3093

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {(2 a b) \int (d \sin (e+f x))^{1+m} \, dx}{d}+\int (d \sin (e+f x))^m \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx \\ & = -\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m)}+\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)}}+\left (a^2+\frac {b^2 (1+m)}{2+m}\right ) \int (d \sin (e+f x))^m \, dx \\ & = -\frac {b^2 \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (2+m)}+\frac {\left (a^2+\frac {b^2 (1+m)}{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) \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.86 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\frac {\sqrt {\cos ^2(e+f x)} (d \sin (e+f x))^m \left (a^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right )+b (1+m) \sin (e+f x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right )+b (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right ) \tan (e+f x)}{f (1+m) (2+m) (3+m)} \]

[In]

Integrate[(d*Sin[e + f*x])^m*(a + b*Sin[e + f*x])^2,x]

[Out]

(Sqrt[Cos[e + f*x]^2]*(d*Sin[e + f*x])^m*(a^2*(6 + 5*m + m^2)*Hypergeometric2F1[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]*Sin[e + f*x]))*Tan[e + f*x])/(f*(1
+ m)*(2 + m)*(3 + m))

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{2}d x\]

[In]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^2,x)

[Out]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^2,x)

Fricas [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*(d*sin(f*x + e))^m, x)

Sympy [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\text {Timed out} \]

[In]

integrate((d*sin(f*x+e))**m*(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^2*(d*sin(f*x + e))^m, x)

Giac [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^2*(d*sin(f*x + e))^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

int((d*sin(e + f*x))^m*(a + b*sin(e + f*x))^2,x)

[Out]

int((d*sin(e + f*x))^m*(a + b*sin(e + f*x))^2, x)