3.2.74 \(\int (c+d x)^m (a+b \sin (e+f x))^3 \, dx\) [174]

Optimal. Leaf size=607 \[ \frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 b^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {3 a^2 b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {3 b^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i 2^{-3-m} a b^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {3 i 2^{-3-m} a b^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-1-m} b^3 e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {3^{-1-m} b^3 e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )}{8 f} \]

[Out]

a^3*(d*x+c)^(1+m)/d/(1+m)+3/2*a*b^2*(d*x+c)^(1+m)/d/(1+m)-3/2*a^2*b*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)-3/8*b^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)-3/2*a^2*b*(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)-3/8*b^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*b^2*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*b^2*(d*x+c)^m*GAMMA(1+m,2*I*f*(d*x+c)/d)/e
xp(2*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)+1/8*3^(-1-m)*b^3*exp(3*I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-3*I*f*(d*x+c)
/d)/f/((-I*f*(d*x+c)/d)^m)+1/8*3^(-1-m)*b^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 = 0.54, antiderivative size = 607, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3398, 3389, 2212, 3393, 3388} \begin {gather*} -\frac {3 a^2 b 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 )}{2 f}-\frac {3 a^2 b 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 )}{2 f}+\frac {3 i a b^2 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 {3 i a b^2 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 {3 b^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 {b^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 {3 b^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 {b^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 {a^3 (c+d x)^{m+1}}{d (m+1)}+\frac {3 a b^2 (c+d x)^{m+1}}{2 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(c + d*x)^(1 + m))/(d*(1 + m)) + (3*a*b^2*(c + d*x)^(1 + m))/(2*d*(1 + m)) - (3*a^2*b*E^(I*(e - (c*f)/d))
*(c + d*x)^m*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(2*f*(((-I)*f*(c + d*x))/d)^m) - (3*b^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) - (3*a^2*b*(c + d*x)^m*Gamma[1 +
m, (I*f*(c + d*x))/d])/(2*E^(I*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) - (3*b^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*b^2*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*b^2*(
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
)*b^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)*b^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 3398

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

Rubi steps

\begin {align*} \int (c+d x)^m (a+b \sin (e+f x))^3 \, dx &=\int \left (a^3 (c+d x)^m+3 a^2 b (c+d x)^m \sin (e+f x)+3 a b^2 (c+d x)^m \sin ^2(e+f x)+b^3 (c+d x)^m \sin ^3(e+f x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\left (3 a^2 b\right ) \int (c+d x)^m \sin (e+f x) \, dx+\left (3 a b^2\right ) \int (c+d x)^m \sin ^2(e+f x) \, dx+b^3 \int (c+d x)^m \sin ^3(e+f x) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {1}{2} \left (3 i a^2 b\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac {1}{2} \left (3 i a^2 b\right ) \int e^{i (e+f x)} (c+d x)^m \, dx+\left (3 a b^2\right ) \int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cos (2 e+2 f x)\right ) \, dx+b^3 \int \left (\frac {3}{4} (c+d x)^m \sin (e+f x)-\frac {1}{4} (c+d x)^m \sin (3 e+3 f x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b 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 )}{2 f}-\frac {3 a^2 b 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 )}{2 f}-\frac {1}{2} \left (3 a b^2\right ) \int (c+d x)^m \cos (2 e+2 f x) \, dx-\frac {1}{4} b^3 \int (c+d x)^m \sin (3 e+3 f x) \, dx+\frac {1}{4} \left (3 b^3\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b 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 )}{2 f}-\frac {3 a^2 b 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 )}{2 f}-\frac {1}{4} \left (3 a b^2\right ) \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a b^2\right ) \int e^{i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{8} \left (i b^3\right ) \int e^{-i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (i b^3\right ) \int e^{i (3 e+3 f x)} (c+d x)^m \, dx+\frac {1}{8} \left (3 i b^3\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac {1}{8} \left (3 i b^3\right ) \int e^{i (e+f x)} (c+d x)^m \, dx\\ &=\frac {a^3 (c+d x)^{1+m}}{d (1+m)}+\frac {3 a b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {3 a^2 b 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 )}{2 f}-\frac {3 b^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 a^2 b 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 )}{2 f}-\frac {3 b^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 b^2 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 b^2 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} b^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} b^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 = 11.85, size = 552, normalized size = 0.91 \begin {gather*} \frac {i (c+d x)^m \left (-24 i a^3 f (c+d x)-36 i a b^2 f (c+d x)+36 i a^2 b d e^{i \left (e-\frac {c f}{d}\right )} (1+m) \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )+9 i b^3 d e^{i \left (e-\frac {c f}{d}\right )} (1+m) \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )+36 i a^2 b d e^{-i \left (e-\frac {c f}{d}\right )} (1+m) \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )+9 i b^3 d e^{-i \left (e-\frac {c f}{d}\right )} (1+m) \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )+9\ 2^{-m} a b^2 d e^{2 i \left (e-\frac {c f}{d}\right )} (1+m) \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )-9\ 2^{-m} a b^2 d e^{-2 i \left (e-\frac {c f}{d}\right )} (1+m) \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )-i 3^{-m} b^3 d e^{3 i \left (e-\frac {c f}{d}\right )} (1+m) \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i f (c+d x)}{d}\right )-i 3^{-m} b^3 d e^{-3 i \left (e-\frac {c f}{d}\right )} (1+m) \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i f (c+d x)}{d}\right )\right )}{24 d f (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/24)*(c + d*x)^m*((-24*I)*a^3*f*(c + d*x) - (36*I)*a*b^2*f*(c + d*x) + ((36*I)*a^2*b*d*E^(I*(e - (c*f)/d))*
(1 + m)*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(((-I)*f*(c + d*x))/d)^m + ((9*I)*b^3*d*E^(I*(e - (c*f)/d))*(1 + m
)*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(((-I)*f*(c + d*x))/d)^m + ((36*I)*a^2*b*d*(1 + m)*Gamma[1 + m, (I*f*(c
+ d*x))/d])/(E^(I*(e - (c*f)/d))*((I*f*(c + d*x))/d)^m) + ((9*I)*b^3*d*(1 + m)*Gamma[1 + m, (I*f*(c + d*x))/d]
)/(E^(I*(e - (c*f)/d))*((I*f*(c + d*x))/d)^m) + (9*a*b^2*d*E^((2*I)*(e - (c*f)/d))*(1 + m)*Gamma[1 + m, ((-2*I
)*f*(c + d*x))/d])/(2^m*(((-I)*f*(c + d*x))/d)^m) - (9*a*b^2*d*(1 + m)*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(2
^m*E^((2*I)*(e - (c*f)/d))*((I*f*(c + d*x))/d)^m) - (I*b^3*d*E^((3*I)*(e - (c*f)/d))*(1 + m)*Gamma[1 + m, ((-3
*I)*f*(c + d*x))/d])/(3^m*(((-I)*f*(c + d*x))/d)^m) - (I*b^3*d*(1 + m)*Gamma[1 + m, ((3*I)*f*(c + d*x))/d])/(3
^m*E^((3*I)*(e - (c*f)/d))*((I*f*(c + d*x))/d)^m)))/(d*f*(1 + m))

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.12, size = 442, normalized size = 0.73 \begin {gather*} -\frac {9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} d m + {\left (4 \, a^{2} b + b^{3}\right )} 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 b^{2} d m - i \, a b^{2} 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 (b^{3} d m + b^{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 ) + 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} d m + {\left (4 \, a^{2} b + b^{3}\right )} 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 b^{2} d m + i \, a b^{2} 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 (b^{3} d m + b^{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 ) - 12 \, {\left ({\left (2 \, a^{3} + 3 \, a b^{2}\right )} d f x + {\left (2 \, a^{3} + 3 \, a b^{2}\right )} 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+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/24*(9*((4*a^2*b + b^3)*d*m + (4*a^2*b + b^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*b^2*d*m - I*a*b^2*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) - (b^3*d*m + b^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) + 9*((4*a^2*b + b^3)*d*m + (4*a^2*b + b^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*b^2*d*m + I*a*b^2*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) - (b^3*d*m + b^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) - 12*((2*a^3 + 3*a*b^2)*d*f*x + (2*a^3 + 3*a*b^2)*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*} \int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \left (c + d x\right )^{m}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(e + f*x))**3*(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+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

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

[Out]

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

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