3.146 \(\int (c+d x)^m (a+a \sin (e+f x))^3 \, dx\)

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

[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 = 0.61, 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 = {3318, 3312, 3307, 2181, 3308} \[ -\frac {15 a^3 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{8 f}+\frac {3 i a^3 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {a^3 3^{-m-1} e^{3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {3 i f (c+d x)}{d}\right )}{8 f}-\frac {15 a^3 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{8 f}-\frac {3 i a^3 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {a^3 3^{-m-1} e^{-3 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {3 i f (c+d x)}{d}\right )}{8 f}+\frac {5 a^3 (c+d x)^{m+1}}{2 d (m+1)} \]

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 2181

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

Rule 3307

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 3308

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 3312

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 3318

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*(e + (Pi*a)/(2*b)))/2 + (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 = 0.90, size = 376, normalized size = 0.84 \[ \frac {1}{24} a^3 (c+d x)^m \left (-\frac {45 e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}+\frac {9 i 2^{-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-m} e^{3 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i f (c+d x)}{d}\right )}{f}-\frac {45 e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{f}-\frac {9 i 2^{-m} e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3^{-m} e^{-3 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i f (c+d x)}{d}\right )}{f}+\frac {60 (c+d x)}{d (m+1)}\right ) \]

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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fricas [A]  time = 0.79, size = 378, normalized size = 0.84 \[ \frac {{\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {3 i \, f}{d}\right ) + 3 i \, d e - 3 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {3 i \, d f x + 3 i \, c f}{d}\right ) + {\left (-9 i \, a^{3} d m - 9 i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, d f x + 2 i \, c f}{d}\right ) - 45 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - 45 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) + {\left (9 i \, a^{3} d m + 9 i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, d f x - 2 i \, c f}{d}\right ) + {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {3 i \, f}{d}\right ) - 3 i \, d e + 3 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-3 i \, d f x - 3 i \, c f}{d}\right ) + 60 \, {\left (a^{3} d f x + a^{3} c f\right )} {\left (d x + c\right )}^{m}}{24 \, {\left (d f m + d f\right )}} \]

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*((a^3*d*m + a^3*d)*e^(-(d*m*log(3*I*f/d) + 3*I*d*e - 3*I*c*f)/d)*gamma(m + 1, (3*I*d*f*x + 3*I*c*f)/d) +
(-9*I*a^3*d*m - 9*I*a^3*d)*e^(-(d*m*log(2*I*f/d) + 2*I*d*e - 2*I*c*f)/d)*gamma(m + 1, (2*I*d*f*x + 2*I*c*f)/d)
- 45*(a^3*d*m + a^3*d)*e^(-(d*m*log(I*f/d) + I*d*e - I*c*f)/d)*gamma(m + 1, (I*d*f*x + I*c*f)/d) - 45*(a^3*d*
m + a^3*d)*e^(-(d*m*log(-I*f/d) - I*d*e + I*c*f)/d)*gamma(m + 1, (-I*d*f*x - I*c*f)/d) + (9*I*a^3*d*m + 9*I*a^
3*d)*e^(-(d*m*log(-2*I*f/d) - 2*I*d*e + 2*I*c*f)/d)*gamma(m + 1, (-2*I*d*f*x - 2*I*c*f)/d) + (a^3*d*m + a^3*d)
*e^(-(d*m*log(-3*I*f/d) - 3*I*d*e + 3*I*c*f)/d)*gamma(m + 1, (-3*I*d*f*x - 3*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (d x + c\right )}^{m}\,{d x} \]

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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maple [F]  time = 0.43, 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 \[ \frac {{\left (d x + c\right )}^{m + 1} a^{3}}{d {\left (m + 1\right )}} + \frac {6 \, a^{3} e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )} - 6 \, {\left (a^{3} d m + a^{3} d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} - {\left (a^{3} d m + a^{3} d\right )} \int {\left (d x + c\right )}^{m} \sin \left (3 \, f x + 3 \, e\right )\,{d x} + 15 \, {\left (a^{3} d m + a^{3} d\right )} \int {\left (d x + c\right )}^{m} \sin \left (f x + e\right )\,{d x}}{4 \, {\left (d m + d\right )}} \]

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

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

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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