3.1.99 \(\int (e x)^m (a+b \sin (c+d x^3))^2 \, dx\) [99]

Optimal. Leaf size=285 \[ \frac {\left (2 a^2+b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i a b e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-i d x^3\right )}{3 e}-\frac {i a b e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},i d x^3\right )}{3 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )}{3 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )}{3 e} \]

[Out]

1/2*(2*a^2+b^2)*(e*x)^(1+m)/e/(1+m)+1/3*I*a*b*exp(I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-I*
d*x^3)/e-1/3*I*a*b*(e*x)^(1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,I*d*x^3)/e/exp(I*c)+1/3*2^(-7/3-1/3*m)*b
^2*exp(2*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-2*I*d*x^3)/e+1/3*2^(-7/3-1/3*m)*b^2*(e*x)^(
1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,2*I*d*x^3)/e/exp(2*I*c)

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Rubi [A]
time = 0.16, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3484, 6, 3471, 2250, 3470} \begin {gather*} \frac {i a b e^{i c} \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 )}{3 e}-\frac {i a b e^{-i c} \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 )}{3 e}+\frac {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 )}{3 e}+\frac {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 )}{3 e}+\frac {\left (2 a^2+b^2\right ) (e x)^{m+1}}{2 e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*(a + b*Sin[c + d*x^3])^2,x]

[Out]

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

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 2250

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

Rule 3470

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]

Rule 3471

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 3484

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]

Rubi steps

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

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Mathematica [A]
time = 5.22, size = 556, normalized size = 1.95 \begin {gather*} \frac {2^{\frac {1}{3} (-7-m)} x (e x)^m \left (d^2 x^6\right )^{\frac {1}{3} (-1-m)} \left (3\ 2^{\frac {7+m}{3}} a^2 \left (d^2 x^6\right )^{\frac {1+m}{3}}+3\ 2^{\frac {4+m}{3}} b^2 \left (d^2 x^6\right )^{\frac {1+m}{3}}+b^2 \left (i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )+b^2 m \left (i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )+b^2 \left (-i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},2 i d x^3\right )+b^2 m \left (-i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},2 i d x^3\right )-i 2^{\frac {7+m}{3}} a b (1+m) \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},i d x^3\right ) (\cos (c)-i \sin (c))+i 2^{\frac {7+m}{3}} a b (1+m) \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-i d x^3\right ) (\cos (c)+i \sin (c))+i b^2 \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right ) \sin (2 c)+i b^2 m \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right ) \sin (2 c)-i b^2 \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right ) \sin (2 c)-i b^2 m \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right ) \sin (2 c)\right )}{3 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*(a + b*Sin[c + d*x^3])^2,x]

[Out]

(2^((-7 - m)/3)*x*(e*x)^m*(d^2*x^6)^((-1 - m)/3)*(3*2^((7 + m)/3)*a^2*(d^2*x^6)^((1 + m)/3) + 3*2^((4 + m)/3)*
b^2*(d^2*x^6)^((1 + m)/3) + b^2*(I*d*x^3)^((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (-2*I)*d*x^3] + b^2*m*(I*d*x^3
)^((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (-2*I)*d*x^3] + b^2*((-I)*d*x^3)^((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3,
(2*I)*d*x^3] + b^2*m*((-I)*d*x^3)^((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (2*I)*d*x^3] - I*2^((7 + m)/3)*a*b*(1
+ m)*((-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + I*2^((7 + m)/3)*a*b*(1 + m)*(I*
d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) + I*b^2*(I*d*x^3)^((1 + m)/3)*Gamma[(1 + m
)/3, (-2*I)*d*x^3]*Sin[2*c] + I*b^2*m*(I*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (-2*I)*d*x^3]*Sin[2*c] - I*b^2*((
-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (2*I)*d*x^3]*Sin[2*c] - I*b^2*m*((-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)
/3, (2*I)*d*x^3]*Sin[2*c]))/(3*(1 + m))

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(a+b*sin(d*x^3+c))^2,x)

[Out]

int((e*x)^m*(a+b*sin(d*x^3+c))^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((e*x)^m*(a+b*sin(d*x^3+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.13, size = 187, normalized size = 0.66 \begin {gather*} \frac {12 \, {\left (2 \, a^{2} + b^{2}\right )} \left (x e\right )^{m} d x - i \, {\left (b^{2} m + b^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (2 i \, d e^{\left (-3\right )}\right ) - 2 i \, c + 2\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 8 \, {\left (a b m + a b\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (i \, d e^{\left (-3\right )}\right ) - i \, c + 2\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 8 \, {\left (a b m + a b\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-i \, d e^{\left (-3\right )}\right ) + i \, c + 2\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) + i \, {\left (b^{2} m + b^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-2 i \, d e^{\left (-3\right )}\right ) + 2 i \, c + 2\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right )}{24 \, {\left (d m + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(a+b*sin(d*x^3+c))^2,x, algorithm="fricas")

[Out]

1/24*(12*(2*a^2 + b^2)*(x*e)^m*d*x - I*(b^2*m + b^2)*e^(-1/3*(m - 2)*log(2*I*d*e^(-3)) - 2*I*c + 2)*gamma(1/3*
m + 1/3, 2*I*d*x^3) - 8*(a*b*m + a*b)*e^(-1/3*(m - 2)*log(I*d*e^(-3)) - I*c + 2)*gamma(1/3*m + 1/3, I*d*x^3) -
8*(a*b*m + a*b)*e^(-1/3*(m - 2)*log(-I*d*e^(-3)) + I*c + 2)*gamma(1/3*m + 1/3, -I*d*x^3) + I*(b^2*m + b^2)*e^
(-1/3*(m - 2)*log(-2*I*d*e^(-3)) + 2*I*c + 2)*gamma(1/3*m + 1/3, -2*I*d*x^3))/(d*m + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(a+b*sin(d*x**3+c))**2,x)

[Out]

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

[Out]

integrate((b*sin(d*x^3 + c) + a)^2*(x*e)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (e\,x\right )}^m\,{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(a + b*sin(c + d*x^3))^2,x)

[Out]

int((e*x)^m*(a + b*sin(c + d*x^3))^2, x)

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