3.2.46 \(\int (c+d x)^m (a+a \sin (e+f x))^3 \, dx\) [146]

Optimal. Leaf size=449 \[ \frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {15 a^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {15 a^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i 2^{-3-m} a^3 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i 2^{-3-m} a^3 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-1-m} a^3 e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {3^{-1-m} a^3 e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )}{8 f} \]

[Out]

5/2*a^3*(d*x+c)^(1+m)/d/(1+m)-15/8*a^3*exp(I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-I*f*(d*x+c)/d)/f/((-I*f*(d*x+c)/d
)^m)-15/8*a^3*(d*x+c)^m*GAMMA(1+m,I*f*(d*x+c)/d)/exp(I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)+3*I*2^(-3-m)*a^3*exp(2
*I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-2*I*f*(d*x+c)/d)/f/((-I*f*(d*x+c)/d)^m)-3*I*2^(-3-m)*a^3*(d*x+c)^m*GAMMA(1+
m,2*I*f*(d*x+c)/d)/exp(2*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)+1/8*3^(-1-m)*a^3*exp(3*I*(e-c*f/d))*(d*x+c)^m*GAMM
A(1+m,-3*I*f*(d*x+c)/d)/f/((-I*f*(d*x+c)/d)^m)+1/8*3^(-1-m)*a^3*(d*x+c)^m*GAMMA(1+m,3*I*f*(d*x+c)/d)/exp(3*I*(
e-c*f/d))/f/((I*f*(d*x+c)/d)^m)

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Rubi [A]
time = 0.41, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3399, 3393, 3388, 2212, 3389} \begin {gather*} -\frac {15 a^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i a^3 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {a^3 3^{-m-1} e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {3 i f (c+d x)}{d}\right )}{8 f}-\frac {15 a^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {3 i a^3 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {a^3 3^{-m-1} e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {5 a^3 (c+d x)^{m+1}}{2 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*(a + a*Sin[e + f*x])^3,x]

[Out]

(5*a^3*(c + d*x)^(1 + m))/(2*d*(1 + m)) - (15*a^3*E^(I*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*f*(c + d*
x))/d])/(8*f*(((-I)*f*(c + d*x))/d)^m) - (15*a^3*(c + d*x)^m*Gamma[1 + m, (I*f*(c + d*x))/d])/(8*E^(I*(e - (c*
f)/d))*f*((I*f*(c + d*x))/d)^m) + ((3*I)*2^(-3 - m)*a^3*E^((2*I)*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*
I)*f*(c + d*x))/d])/(f*(((-I)*f*(c + d*x))/d)^m) - ((3*I)*2^(-3 - m)*a^3*(c + d*x)^m*Gamma[1 + m, ((2*I)*f*(c
+ d*x))/d])/(E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (3^(-1 - m)*a^3*E^((3*I)*(e - (c*f)/d))*(c + d
*x)^m*Gamma[1 + m, ((-3*I)*f*(c + d*x))/d])/(8*f*(((-I)*f*(c + d*x))/d)^m) + (3^(-1 - m)*a^3*(c + d*x)^m*Gamma
[1 + m, ((3*I)*f*(c + d*x))/d])/(8*E^((3*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int (c+d x)^m (a+a \sin (e+f x))^3 \, dx &=\left (8 a^3\right ) \int (c+d x)^m \sin ^6\left (\frac {1}{2} \left (e+\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx\\ &=\left (8 a^3\right ) \int \left (\frac {5}{16} (c+d x)^m-\frac {3}{16} (c+d x)^m \cos (2 e+2 f x)+\frac {15}{32} (c+d x)^m \sin (e+f x)-\frac {1}{32} (c+d x)^m \sin (3 e+3 f x)\right ) \, dx\\ &=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{4} a^3 \int (c+d x)^m \sin (3 e+3 f x) \, dx-\frac {1}{2} \left (3 a^3\right ) \int (c+d x)^m \cos (2 e+2 f x) \, dx+\frac {1}{4} \left (15 a^3\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{8} \left (i a^3\right ) \int e^{-i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (i a^3\right ) \int e^{i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (15 i a^3\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac {1}{8} \left (15 i a^3\right ) \int e^{i (e+f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a^3\right ) \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a^3\right ) \int e^{i (2 e+2 f x)} (c+d x)^m \, dx\\ &=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {15 a^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {15 a^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i 2^{-3-m} a^3 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i 2^{-3-m} a^3 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-1-m} a^3 e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {3^{-1-m} a^3 e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )}{8 f}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 376, normalized size = 0.84 \begin {gather*} \frac {1}{24} a^3 (c+d x)^m \left (\frac {60 (c+d x)}{d (1+m)}-\frac {45 e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {45 e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {9 i 2^{-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {9 i 2^{-m} e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-m} e^{3 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )}{f}+\frac {3^{-m} e^{-3 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )}{f}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*(a + a*Sin[e + f*x])^3,x]

[Out]

(a^3*(c + d*x)^m*((60*(c + d*x))/(d*(1 + m)) - (45*E^(I*(e - (c*f)/d))*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(f*
(((-I)*f*(c + d*x))/d)^m) - (45*Gamma[1 + m, (I*f*(c + d*x))/d])/(E^(I*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)
+ ((9*I)*E^((2*I)*(e - (c*f)/d))*Gamma[1 + m, ((-2*I)*f*(c + d*x))/d])/(2^m*f*(((-I)*f*(c + d*x))/d)^m) - ((9
*I)*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(2^m*E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (E^((3*I)*(e
- (c*f)/d))*Gamma[1 + m, ((-3*I)*f*(c + d*x))/d])/(3^m*f*(((-I)*f*(c + d*x))/d)^m) + Gamma[1 + m, ((3*I)*f*(c
+ d*x))/d]/(3^m*E^((3*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)))/24

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (a +a \sin \left (f x +e \right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*(a+a*sin(f*x+e))^3,x)

[Out]

int((d*x+c)^m*(a+a*sin(f*x+e))^3,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((d*x+c)^m*(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

(d*x + c)^(m + 1)*a^3/(d*(m + 1)) + 1/4*(6*a^3*e^(m*log(d*x + c) + log(d*x + c)) - 6*(a^3*d*m + a^3*d)*integra
te((d*x + c)^m*cos(2*f*x + 2*e), x) - (a^3*d*m + a^3*d)*integrate((d*x + c)^m*sin(3*f*x + 3*e), x) + 15*(a^3*d
*m + a^3*d)*integrate((d*x + c)^m*sin(f*x + e), x))/(d*m + d)

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Fricas [A]
time = 0.13, size = 392, normalized size = 0.87 \begin {gather*} -\frac {45 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) - i \, c f + i \, d e}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) + 9 \, {\left (-i \, a^{3} d m - i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) + 2 i \, c f - 2 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) - {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {3 i \, f}{d}\right ) + 3 i \, c f - 3 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) + 45 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) + i \, c f - i \, d e}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) + 9 \, {\left (i \, a^{3} d m + i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) - 2 i \, c f + 2 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {3 i \, f}{d}\right ) - 3 i \, c f + 3 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 60 \, {\left (a^{3} d f x + a^{3} c f\right )} {\left (d x + c\right )}^{m}}{24 \, {\left (d f m + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/24*(45*(a^3*d*m + a^3*d)*e^(-(d*m*log(I*f/d) - I*c*f + I*d*e)/d)*gamma(m + 1, (I*d*f*x + I*c*f)/d) + 9*(-I*
a^3*d*m - I*a^3*d)*e^(-(d*m*log(-2*I*f/d) + 2*I*c*f - 2*I*d*e)/d)*gamma(m + 1, -2*(I*d*f*x + I*c*f)/d) - (a^3*
d*m + a^3*d)*e^(-(d*m*log(-3*I*f/d) + 3*I*c*f - 3*I*d*e)/d)*gamma(m + 1, -3*(I*d*f*x + I*c*f)/d) + 45*(a^3*d*m
+ a^3*d)*e^(-(d*m*log(-I*f/d) + I*c*f - I*d*e)/d)*gamma(m + 1, (-I*d*f*x - I*c*f)/d) + 9*(I*a^3*d*m + I*a^3*d
)*e^(-(d*m*log(2*I*f/d) - 2*I*c*f + 2*I*d*e)/d)*gamma(m + 1, -2*(-I*d*f*x - I*c*f)/d) - (a^3*d*m + a^3*d)*e^(-
(d*m*log(3*I*f/d) - 3*I*c*f + 3*I*d*e)/d)*gamma(m + 1, -3*(-I*d*f*x - I*c*f)/d) - 60*(a^3*d*f*x + a^3*c*f)*(d*
x + c)^m)/(d*f*m + d*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \left (c + d x\right )^{m} \sin {\left (e + f x \right )}\, dx + \int 3 \left (c + d x\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m} \sin ^{3}{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m*(a+a*sin(f*x+e))**3,x)

[Out]

a**3*(Integral(3*(c + d*x)**m*sin(e + f*x), x) + Integral(3*(c + d*x)**m*sin(e + f*x)**2, x) + Integral((c + d
*x)**m*sin(e + f*x)**3, x) + Integral((c + d*x)**m, 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((d*x+c)^m*(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^3*(d*x + c)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c+d\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3*(c + d*x)^m,x)

[Out]

int((a + a*sin(e + f*x))^3*(c + d*x)^m, x)

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