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

Optimal. Leaf size=442 \[ \frac{i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-i d x^3\right )}{8 e}-\frac{i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},i d x^3\right )}{8 e}+\frac{a b^2 e^{2 i c} 2^{-\frac{m}{3}-\frac{7}{3}} \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-2 i d x^3\right )}{e}+\frac{a b^2 e^{-2 i c} 2^{-\frac{m}{3}-\frac{7}{3}} \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},2 i d x^3\right )}{e}-\frac{i b^3 e^{3 i c} 3^{-\frac{m}{3}-\frac{4}{3}} \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-3 i d x^3\right )}{8 e}+\frac{i b^3 e^{-3 i c} 3^{-\frac{m}{3}-\frac{4}{3}} \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},3 i d x^3\right )}{8 e}+\frac{a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]

[Out]

(a*(2*a^2 + 3*b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/8)*b*(4*a^2 + b^2)*E^(I*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((
-1 - m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3])/e - ((I/8)*b*(4*a^2 + b^2)*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma
[(1 + m)/3, I*d*x^3])/(e*E^(I*c)) + (2^(-7/3 - m/3)*a*b^2*E^((2*I)*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3)*
Gamma[(1 + m)/3, (-2*I)*d*x^3])/e + (2^(-7/3 - m/3)*a*b^2*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3
, (2*I)*d*x^3])/(e*E^((2*I)*c)) - ((I/8)*3^(-4/3 - m/3)*b^3*E^((3*I)*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3
)*Gamma[(1 + m)/3, (-3*I)*d*x^3])/e + ((I/8)*3^(-4/3 - m/3)*b^3*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1
+ m)/3, (3*I)*d*x^3])/(e*E^((3*I)*c))

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Rubi [A]  time = 0.414381, antiderivative size = 442, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3403, 6, 3390, 2218, 3389} \[ \frac{i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-i d x^3\right )}{8 e}-\frac{i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},i d x^3\right )}{8 e}+\frac{a b^2 e^{2 i c} 2^{-\frac{m}{3}-\frac{7}{3}} \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-2 i d x^3\right )}{e}+\frac{a b^2 e^{-2 i c} 2^{-\frac{m}{3}-\frac{7}{3}} \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},2 i d x^3\right )}{e}-\frac{i b^3 e^{3 i c} 3^{-\frac{m}{3}-\frac{4}{3}} \left (-i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},-3 i d x^3\right )}{8 e}+\frac{i b^3 e^{-3 i c} 3^{-\frac{m}{3}-\frac{4}{3}} \left (i d x^3\right )^{\frac{1}{3} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{3},3 i d x^3\right )}{8 e}+\frac{a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*a^2 + 3*b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/8)*b*(4*a^2 + b^2)*E^(I*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((
-1 - m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3])/e - ((I/8)*b*(4*a^2 + b^2)*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma
[(1 + m)/3, I*d*x^3])/(e*E^(I*c)) + (2^(-7/3 - m/3)*a*b^2*E^((2*I)*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3)*
Gamma[(1 + m)/3, (-2*I)*d*x^3])/e + (2^(-7/3 - m/3)*a*b^2*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3
, (2*I)*d*x^3])/(e*E^((2*I)*c)) - ((I/8)*3^(-4/3 - m/3)*b^3*E^((3*I)*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3
)*Gamma[(1 + m)/3, (-3*I)*d*x^3])/e + ((I/8)*3^(-4/3 - m/3)*b^3*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1
+ m)/3, (3*I)*d*x^3])/(e*E^((3*I)*c))

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 3390

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] + Dist[1/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3389

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] - Dist[I/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps

\begin{align*} \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx &=\int \left (a^3 (e x)^m+\frac{3}{2} a b^2 (e x)^m-\frac{3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+3 a^2 b (e x)^m \sin \left (c+d x^3\right )+\frac{3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac{1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^3+\frac{3 a b^2}{2}\right ) (e x)^m-\frac{3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+3 a^2 b (e x)^m \sin \left (c+d x^3\right )+\frac{3}{4} b^3 (e x)^m \sin \left (c+d x^3\right )-\frac{1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^3+\frac{3 a b^2}{2}\right ) (e x)^m-\frac{3}{2} a b^2 (e x)^m \cos \left (2 c+2 d x^3\right )+\left (3 a^2 b+\frac{3 b^3}{4}\right ) (e x)^m \sin \left (c+d x^3\right )-\frac{1}{4} b^3 (e x)^m \sin \left (3 c+3 d x^3\right )\right ) \, dx\\ &=\frac{a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}-\frac{1}{2} \left (3 a b^2\right ) \int (e x)^m \cos \left (2 c+2 d x^3\right ) \, dx-\frac{1}{4} b^3 \int (e x)^m \sin \left (3 c+3 d x^3\right ) \, dx+\frac{1}{4} \left (3 b \left (4 a^2+b^2\right )\right ) \int (e x)^m \sin \left (c+d x^3\right ) \, dx\\ &=\frac{a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}-\frac{1}{4} \left (3 a b^2\right ) \int e^{-2 i c-2 i d x^3} (e x)^m \, dx-\frac{1}{4} \left (3 a b^2\right ) \int e^{2 i c+2 i d x^3} (e x)^m \, dx-\frac{1}{8} \left (i b^3\right ) \int e^{-3 i c-3 i d x^3} (e x)^m \, dx+\frac{1}{8} \left (i b^3\right ) \int e^{3 i c+3 i d x^3} (e x)^m \, dx+\frac{1}{8} \left (3 i b \left (4 a^2+b^2\right )\right ) \int e^{-i c-i d x^3} (e x)^m \, dx-\frac{1}{8} \left (3 i b \left (4 a^2+b^2\right )\right ) \int e^{i c+i d x^3} (e x)^m \, dx\\ &=\frac{a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac{i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},-i d x^3\right )}{8 e}-\frac{i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},i d x^3\right )}{8 e}+\frac{2^{-\frac{7}{3}-\frac{m}{3}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},-2 i d x^3\right )}{e}+\frac{2^{-\frac{7}{3}-\frac{m}{3}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},2 i d x^3\right )}{e}-\frac{i 3^{-\frac{4}{3}-\frac{m}{3}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},-3 i d x^3\right )}{8 e}+\frac{i 3^{-\frac{4}{3}-\frac{m}{3}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac{1}{3} (-1-m)} \Gamma \left (\frac{1+m}{3},3 i d x^3\right )}{8 e}\\ \end{align*}

Mathematica [A]  time = 12.4702, size = 373, normalized size = 0.84 \[ \frac{1}{24} i x (e x)^m \left (3 b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},-i d x^3\right )-3 b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},i d x^3\right )-3 i a b^2 e^{2 i c} 2^{\frac{2}{3}-\frac{m}{3}} \left (-i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},-2 i d x^3\right )-3 i a b^2 e^{-2 i c} 2^{\frac{2}{3}-\frac{m}{3}} \left (i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},2 i d x^3\right )-b^3 e^{3 i c} 3^{-\frac{m}{3}-\frac{1}{3}} \left (-i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},-3 i d x^3\right )+b^3 e^{-3 i c} 3^{-\frac{m}{3}-\frac{1}{3}} \left (i d x^3\right )^{-\frac{m}{3}-\frac{1}{3}} \text{Gamma}\left (\frac{m+1}{3},3 i d x^3\right )-\frac{12 i a \left (2 a^2+3 b^2\right )}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/24)*x*(e*x)^m*(((-12*I)*a*(2*a^2 + 3*b^2))/(1 + m) + 3*b*(4*a^2 + b^2)*E^(I*c)*((-I)*d*x^3)^(-1/3 - m/3)*Ga
mma[(1 + m)/3, (-I)*d*x^3] - (3*b*(4*a^2 + b^2)*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, I*d*x^3])/E^(I*c) - (3
*I)*2^(2/3 - m/3)*a*b^2*E^((2*I)*c)*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-2*I)*d*x^3] - ((3*I)*2^(2/3 -
m/3)*a*b^2*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (2*I)*d*x^3])/E^((2*I)*c) - 3^(-1/3 - m/3)*b^3*E^((3*I)*c)
*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-3*I)*d*x^3] + (3^(-1/3 - m/3)*b^3*(I*d*x^3)^(-1/3 - m/3)*Gamma[(
1 + m)/3, (3*I)*d*x^3])/E^((3*I)*c))

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Maple [F]  time = 0.389, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00216, size = 926, normalized size = 2.1 \begin{align*} \frac{36 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (e x\right )^{m} d x +{\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (\frac{3 i \, d}{e^{3}}\right ) - 3 i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, 3 i \, d x^{3}\right ) +{\left (-9 i \, a b^{2} e^{2} m - 9 i \, a b^{2} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (\frac{2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, 2 i \, d x^{3}\right ) - 9 \,{\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m +{\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (\frac{i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, i \, d x^{3}\right ) - 9 \,{\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m +{\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (-\frac{i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, -i \, d x^{3}\right ) +{\left (9 i \, a b^{2} e^{2} m + 9 i \, a b^{2} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (-\frac{2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, -2 i \, d x^{3}\right ) +{\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac{1}{3} \,{\left (m - 2\right )} \log \left (-\frac{3 i \, d}{e^{3}}\right ) + 3 i \, c\right )} \Gamma \left (\frac{1}{3} \, m + \frac{1}{3}, -3 i \, d x^{3}\right )}{72 \,{\left (d m + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(36*(2*a^3 + 3*a*b^2)*(e*x)^m*d*x + (b^3*e^2*m + b^3*e^2)*e^(-1/3*(m - 2)*log(3*I*d/e^3) - 3*I*c)*gamma(1
/3*m + 1/3, 3*I*d*x^3) + (-9*I*a*b^2*e^2*m - 9*I*a*b^2*e^2)*e^(-1/3*(m - 2)*log(2*I*d/e^3) - 2*I*c)*gamma(1/3*
m + 1/3, 2*I*d*x^3) - 9*((4*a^2*b + b^3)*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*(m - 2)*log(I*d/e^3) - I*c)*gamm
a(1/3*m + 1/3, I*d*x^3) - 9*((4*a^2*b + b^3)*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*(m - 2)*log(-I*d/e^3) + I*c)
*gamma(1/3*m + 1/3, -I*d*x^3) + (9*I*a*b^2*e^2*m + 9*I*a*b^2*e^2)*e^(-1/3*(m - 2)*log(-2*I*d/e^3) + 2*I*c)*gam
ma(1/3*m + 1/3, -2*I*d*x^3) + (b^3*e^2*m + b^3*e^2)*e^(-1/3*(m - 2)*log(-3*I*d/e^3) + 3*I*c)*gamma(1/3*m + 1/3
, -3*I*d*x^3))/(d*m + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{3} \left (e x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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