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

Optimal. Leaf size=194 \[ \frac{\left (a^2 (m+2)+b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt{\cos ^2(e+f x)}}+\frac{2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt{\cos ^2(e+f x)}}-\frac{b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)} \]

[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])

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Rubi [A]  time = 0.134654, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2789, 2643, 3014} \[ \frac{\left (a^2 (m+2)+b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt{\cos ^2(e+f x)}}+\frac{2 a b \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt{\cos ^2(e+f x)}}-\frac{b^2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2)} \]

Antiderivative was successfully verified.

[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 2789

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3014

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*} \int (d \sin (e+f x))^m (a+b \sin (e+f x))^2 \, dx &=\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) \, _2F_1\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) \, _2F_1\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) \, _2F_1\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]  time = 0.305454, size = 144, normalized size = 0.74 \[ -\frac{\cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-m-1)} (d \sin (e+f x))^m \left (a \left (a \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3}{2};\cos ^2(e+f x)\right )+2 b \sqrt{\sin ^2(e+f x)} \, _2F_1\left (\frac{1}{2},-\frac{m}{2};\frac{3}{2};\cos ^2(e+f x)\right )\right )+b^2 \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-1);\frac{3}{2};\cos ^2(e+f x)\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*(d*Sin[e + f*x])^m*(Sin[e + f*x]^2)^((-1 - m)/2)*(b^2*Hypergeometric2F1[1/2, (-1 - m)/2, 3/2,
Cos[e + f*x]^2]*Sin[e + f*x] + a*(a*Hypergeometric2F1[1/2, (1 - m)/2, 3/2, Cos[e + f*x]^2]*Sin[e + f*x] + 2*b*
Hypergeometric2F1[1/2, -m/2, 3/2, Cos[e + f*x]^2]*Sqrt[Sin[e + f*x]^2])))/f)

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Maple [F]  time = 3.307, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \left ( a+b\sin \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\right )} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)