3.17 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=184 \[ -\frac{2^{m+\frac{1}{2}} \left (A \left (m^2+3 m+2\right )+B m (m+2)+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{(C-B (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

((C - B*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (2^(1/2 + m)*(B*m*(2 + m) + C*(1 +
m + m^2) + A*(2 + 3*m + m^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Si
n[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 +
m))/(a*f*(2 + m))

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Rubi [A]  time = 0.231302, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3023, 2751, 2652, 2651} \[ -\frac{2^{m+\frac{1}{2}} \left (A \left (m^2+3 m+2\right )+B m (m+2)+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{(C-B (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

((C - B*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (2^(1/2 + m)*(B*m*(2 + m) + C*(1 +
m + m^2) + A*(2 + 3*m + m^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Si
n[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 +
m))/(a*f*(2 + m))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !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 \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac{C \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 (a (C (1+m)+A (2+m))-a (C-B (2+m)) \sin (e+f x)) \, dx}{a (2+m)}\\ &=\frac{(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (\left (B m (2+m)+C \left (1+m+m^2\right )+A \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{(C-B (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 (B m (2+m)+C \left (1+m+m^2\right )+A \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{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}\\ \end{align*}

Mathematica [C]  time = 5.08781, size = 525, normalized size = 2.85 \[ -\frac{\sin ^{-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (\frac{4 \sqrt{2} A \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt{1-\sin (e+f x)}}+\frac{\sqrt [4]{-1} B 4^{-m} e^{-\frac{3}{2} i (e+f x)} \left (-(-1)^{3/4} e^{-\frac{1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m+1} \left ((m-1) e^{2 i (e+f x)} \, _2F_1\left (1,m;-m;-i e^{-i (e+f x)}\right )-(m+1) \, _2F_1\left (1,m+2;2-m;-i e^{-i (e+f x)}\right )\right )}{m^2-1}+\frac{C 2^{-2 m-1} e^{-3 i (e+f x)} \left (1-i e^{i (e+f x)}\right ) \left (-(-1)^{3/4} e^{-\frac{1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m} \left ((m-2) e^{4 i (e+f x)} \, _2F_1\left (1,m-1;-m-1;-i e^{-i (e+f x)}\right )+(m+2) \, _2F_1\left (1,m+3;3-m;-i e^{-i (e+f x)}\right )\right )}{m^2-4}+\frac{2 \sqrt{2} C \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt{1-\sin (e+f x)}}\right )}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

-((a*(1 + Sin[e + f*x]))^m*(((-1)^(1/4)*B*(-(((-1)^(3/4)*(I + E^(I*(e + f*x))))/E^((I/2)*(e + f*x))))^(1 + 2*m
)*(E^((2*I)*(e + f*x))*(-1 + m)*Hypergeometric2F1[1, m, -m, (-I)/E^(I*(e + f*x))] - (1 + m)*Hypergeometric2F1[
1, 2 + m, 2 - m, (-I)/E^(I*(e + f*x))]))/(4^m*E^(((3*I)/2)*(e + f*x))*(-1 + m^2)) + (2^(-1 - 2*m)*C*(1 - I*E^(
I*(e + f*x)))*(-(((-1)^(3/4)*(I + E^(I*(e + f*x))))/E^((I/2)*(e + f*x))))^(2*m)*(E^((4*I)*(e + f*x))*(-2 + m)*
Hypergeometric2F1[1, -1 + m, -1 - m, (-I)/E^(I*(e + f*x))] + (2 + m)*Hypergeometric2F1[1, 3 + m, 3 - m, (-I)/E
^(I*(e + f*x))]))/(E^((3*I)*(e + f*x))*(-4 + m^2)) + (4*Sqrt[2]*A*Cos[(2*e - Pi + 2*f*x)/4]^(1 + 2*m)*Hypergeo
metric2F1[1/2, 1/2 + m, 3/2 + m, Sin[(2*e + Pi + 2*f*x)/4]^2]*Sin[(2*e - Pi + 2*f*x)/4])/((1 + 2*m)*Sqrt[1 - S
in[e + f*x]]) + (2*Sqrt[2]*C*Cos[(2*e - Pi + 2*f*x)/4]^(1 + 2*m)*Hypergeometric2F1[1/2, 1/2 + m, 3/2 + m, Sin[
(2*e + Pi + 2*f*x)/4]^2]*Sin[(2*e - Pi + 2*f*x)/4])/((1 + 2*m)*Sqrt[1 - Sin[e + f*x]])))/(2*f*Sin[(2*e + Pi +
2*f*x)/4]^(2*m))

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Maple [F]  time = 1.59, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\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*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(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 (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\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*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(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 (A + B \sin{\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x) + C*sin(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\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*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m, x)