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

Optimal. Leaf size=171 \[ -\frac {2^{m+\frac {1}{2}} \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}-\frac {C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

C*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(m^2+3*m+2)-2^(1/2+m)*(C*(m^2+m+1)+A*(m^2+3*m+2))*cos(f*x+e)*hypergeom([1/2,
1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m/f/(m^2+3*m+2)-C*cos(f*x+e)*(a+a*s
in(f*x+e))^(1+m)/a/f/(2+m)

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Rubi [A]  time = 0.19, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3024, 2751, 2652, 2651} \[ -\frac {2^{m+\frac {1}{2}} \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}-\frac {C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(2 + 3*m + m^2)) - (2^(1/2 + 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 + Sin[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 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]

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 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 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rubi steps

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

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Mathematica [C]  time = 2.54, size = 385, normalized size = 2.25 \[ -\frac {\sin ^{-2 m}\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (\frac {4 \sqrt {2} A \sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};\sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt {1-\sin (e+f x)}}+\frac {C 2^{-2 m-1} e^{-3 i (e+f x)} \left (1-i e^{i (e+f x)}\right ) \left (-(-1)^{3/4} e^{-\frac {1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m} \left ((m-2) e^{4 i (e+f x)} \, _2F_1\left (1,m-1;-m-1;-i e^{-i (e+f x)}\right )+(m+2) \, _2F_1\left (1,m+3;3-m;-i e^{-i (e+f x)}\right )\right )}{m^2-4}+\frac {2 \sqrt {2} C \sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};\sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt {1-\sin (e+f x)}}\right )}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*((a*(1 + Sin[e + f*x]))^m*((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)*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*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]])))/(f*Sin[(2*e + Pi + 2*f*x)/4]^(2*m))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - A - C\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.53, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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