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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.23, 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 = {3403, 6, 3390, 2218, 3389} \[ \frac {i a b e^{i c} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},-i d x^3\right )}{3 e}-\frac {i a b e^{-i c} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},i d x^3\right )}{3 e}+\frac {b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},-2 i d x^3\right )}{3 e}+\frac {b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{3},2 i d x^3\right )}{3 e}+\frac {\left (2 a^2+b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]

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 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]

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 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]

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*}

________________________________________________________________________________________

Mathematica [A]  time = 6.67, size = 556, normalized size = 1.95 \[ \frac {2^{\frac {1}{3} (-m-7)} x \left (d^2 x^6\right )^{\frac {1}{3} (-m-1)} (e x)^m \left (3 a^2 2^{\frac {m+7}{3}} \left (d^2 x^6\right )^{\frac {m+1}{3}}-i a b 2^{\frac {m+7}{3}} (m+1) (\cos (c)-i \sin (c)) \left (-i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},i d x^3\right )+i a b 2^{\frac {m+7}{3}} (m+1) (\cos (c)+i \sin (c)) \left (i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},-i d x^3\right )+b^2 \cos (2 c) \left (-i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )+b^2 m \cos (2 c) \left (-i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )+b^2 \cos (2 c) \left (i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )+b^2 m \cos (2 c) \left (i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )-i b^2 \sin (2 c) \left (-i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )-i b^2 m \sin (2 c) \left (-i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )+i b^2 \sin (2 c) \left (i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )+i b^2 m \sin (2 c) \left (i d x^3\right )^{\frac {m+1}{3}} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )+3 b^2 2^{\frac {m+4}{3}} \left (d^2 x^6\right )^{\frac {m+1}{3}}\right )}{3 (m+1)} \]

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))

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 214, normalized size = 0.75 \[ \frac {12 \, {\left (2 \, a^{2} + b^{2}\right )} \left (e x\right )^{m} d x + {\left (-i \, b^{2} e^{2} m - i \, b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 8 \, {\left (a b e^{2} m + a b e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 8 \, {\left (a b e^{2} m + a b e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) + {\left (i \, b^{2} e^{2} m + i \, b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right )}{24 \, {\left (d m + d\right )}} \]

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

________________________________________________________________________________________

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

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*(e*x)^m, x)

________________________________________________________________________________________

maple [F]  time = 0.70, 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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\left (e x\right )^{m + 1} a^{2}}{e {\left (m + 1\right )}} + \frac {b^{2} e^{m} x x^{m} - {\left (b^{2} e^{m} m + b^{2} e^{m}\right )} \int x^{m} \cos \left (2 \, d x^{3} + 2 \, c\right )\,{d x} + 4 \, {\left (a b e^{m} m + a b e^{m}\right )} \int x^{m} \sin \left (d x^{3} + c\right )\,{d x}}{2 \, {\left (m + 1\right )}} \]

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]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________