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

Optimal. Leaf size=139 \[ \frac{a \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) \sqrt{\cos ^2(e+f x)}}+\frac{b \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt{\cos ^2(e+f x)}} \]

[Out]

(a*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]) + (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 = 0.0722351, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2748, 2643} \[ \frac{a \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) \sqrt{\cos ^2(e+f x)}}+\frac{b \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*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]) + (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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], 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]

Rubi steps

\begin{align*} \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx &=a \int (d \sin (e+f x))^m \, dx+\frac{b \int (d \sin (e+f x))^{1+m} \, dx}{d}\\ &=\frac{a \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{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 = 0.156493, size = 111, normalized size = 0.8 \[ \frac{\sqrt{\cos ^2(e+f x)} \tan (e+f x) (d \sin (e+f x))^m \left (a (m+2) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(e+f x)\right )+b (m+1) \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )\right )}{f (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[e + f*x]^2]*(d*Sin[e + f*x])^m*(a*(2 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^
2] + b*(1 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*Sin[e + f*x])*Tan[e + f*x])/(f*(1
+ m)*(2 + m))

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Maple [F]  time = 1.076, 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 ) \, 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)),x)

[Out]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),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 )} \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)),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)*(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 \sin \left (f x + e\right ) + a\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)),x, algorithm="fricas")

[Out]

integral((b*sin(f*x + e) + a)*(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)),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 )} \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)),x, algorithm="giac")

[Out]

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