3.1.13 \(\int (a+a \sin (e+f x))^m (A+C \sin ^2(e+f x)) \, dx\) [13]

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

[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 = 0.14, 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 = {3103, 2830, 2731, 2730} \begin {gather*} -\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)} \end {gather*}

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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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*S
in[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 3103

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] &&  !Lt
Q[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] Result contains complex when optimal does not.
time = 1.83, size = 385, normalized size = 2.25 \begin {gather*} -\frac {(a (1+\sin (e+f x)))^m \left (\frac {2^{-1-2 m} C 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 (i+e^{i (e+f x)}\right )\right )^{2 m} \left (e^{4 i (e+f x)} (-2+m) \, _2F_1\left (1,-1+m;-1-m;-i e^{-i (e+f x)}\right )+(2+m) \, _2F_1\left (1,3+m;3-m;-i e^{-i (e+f x)}\right )\right )}{-4+m^2}+\frac {4 \sqrt {2} A \cos ^{1+2 m}\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{(1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2 \sqrt {2} C \cos ^{1+2 m}\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{(1+2 m) \sqrt {1-\sin (e+f x)}}\right ) \sin ^{-2 m}\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{2 f} \end {gather*}

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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Maple [F]
time = 0.42, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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