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

Optimal. Leaf size=193 \[ -\frac{2^{m+\frac{1}{2}} \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.268819, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2761, 2751, 2652, 2651} \[ -\frac{2^{m+\frac{1}{2}} \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d
, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx &=-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\int (a+a \sin (e+f x))^m \left (a \left (d^2 (1+m)+c^2 (2+m)\right )-a d (d-2 c (2+m)) \sin (e+f x)\right ) \, dx}{a (2+m)}\\ &=\frac{d (d-2 c (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{d (d-2 c (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (\left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{d (d-2 c (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{2^{\frac{1}{2}+m} \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}\\ \end{align*}

Mathematica [B]  time = 103.011, size = 1774, normalized size = 9.19 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(Cos[(-e + Pi/2 - f*x)/2]^2)^(1/2 - m)*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^(-1/2 + m)*(c + d - 2*d*Sin[(-e +
Pi/2 - f*x)/2]^2)^2*(4*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 9/2}, Sin[(-e + Pi/2 - f*x)/2]^
2]*Sin[(-e + Pi/2 - f*x)/2]^2*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^2 + 16*Gamma[3/2 - m]*Hypergeometric2F1
[3/2, 3/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2*(c^2 + c*d*(2 - 3*Sin[(-e + Pi/2 -
f*x)/2]^2) + d^2*(1 - 3*Sin[(-e + Pi/2 - f*x)/2]^2 + 2*Sin[(-e + Pi/2 - f*x)/2]^4)) + 7*Gamma[1/2 - m]*Hyperge
ometric2F1[1/2, 1/2 - m, 7/2, Sin[(-e + Pi/2 - f*x)/2]^2]*(15*c^2 + 10*c*d*(3 - 2*Sin[(-e + Pi/2 - f*x)/2]^2)
+ d^2*(15 - 20*Sin[(-e + Pi/2 - f*x)/2]^2 + 12*Sin[(-e + Pi/2 - f*x)/2]^4)))*(a + a*Sin[e + f*x])^m*Tan[(-e +
Pi/2 - f*x)/2])/(f*(4*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 9/2}, Sin[(-e + Pi/2 - f*x)/2]^2
]*Sin[(-e + Pi/2 - f*x)/2]^2*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^2 + 16*Gamma[3/2 - m]*Hypergeometric2F1[
3/2, 3/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2*(c^2 + c*d*(2 - 3*Sin[(-e + Pi/2 - f
*x)/2]^2) + d^2*(1 - 3*Sin[(-e + Pi/2 - f*x)/2]^2 + 2*Sin[(-e + Pi/2 - f*x)/2]^4)) + 7*Gamma[1/2 - m]*Hypergeo
metric2F1[1/2, 1/2 - m, 7/2, Sin[(-e + Pi/2 - f*x)/2]^2]*(15*c^2 + 10*c*d*(3 - 2*Sin[(-e + Pi/2 - f*x)/2]^2) +
d^2*(15 - 20*Sin[(-e + Pi/2 - f*x)/2]^2 + 12*Sin[(-e + Pi/2 - f*x)/2]^4)) + (2*Sin[(-e + Pi/2 - f*x)/2]^2*(-4
8*d*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 9/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 -
f*x)/2]^2*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2) + 12*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {
1, 9/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^2 - 4*(-3 + 2*m)*Gamma[3/2 - m]*
HypergeometricPFQ[{5/2, 3, 5/2 - m}, {2, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2*(c + d
- 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^2 + 48*d*Gamma[3/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 9/2, Sin[(-e + Pi/2
- f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]*(-3*c*Sin[(-e + Pi/2 - f*x)/2] + d*Sin[(-e + Pi/2 - f*x)/2]*(-3 + 4*Sin[
(-e + Pi/2 - f*x)/2]^2)) + 84*d*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 7/2, Sin[(-e + Pi/2 - f*x)/2]^2
]*(-5*c + d*(-5 + 6*Sin[(-e + Pi/2 - f*x)/2]^2)) + 48*Gamma[3/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 9/2, Sin[
(-e + Pi/2 - f*x)/2]^2]*(c^2 + c*d*(2 - 3*Sin[(-e + Pi/2 - f*x)/2]^2) + d^2*(1 - 3*Sin[(-e + Pi/2 - f*x)/2]^2
+ 2*Sin[(-e + Pi/2 - f*x)/2]^4)) - 8*(-3 + 2*m)*Gamma[3/2 - m]*Hypergeometric2F1[5/2, 5/2 - m, 11/2, Sin[(-e +
Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2*(c^2 + c*d*(2 - 3*Sin[(-e + Pi/2 - f*x)/2]^2) + d^2*(1 - 3*Sin[(
-e + Pi/2 - f*x)/2]^2 + 2*Sin[(-e + Pi/2 - f*x)/2]^4)) + 3*(1/2 - m)*Gamma[1/2 - m]*Hypergeometric2F1[3/2, 3/2
- m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*(15*c^2 + 10*c*d*(3 - 2*Sin[(-e + Pi/2 - f*x)/2]^2) + d^2*(15 - 20*Sin[
(-e + Pi/2 - f*x)/2]^2 + 12*Sin[(-e + Pi/2 - f*x)/2]^4))))/3))

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Maple [F]  time = 2.927, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (c + d \sin{\left (e + f x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(c + d*sin(e + f*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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