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

Optimal. Leaf size=279 \[ \frac{i a b e^{i c} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )}{2 e}-\frac{i a b e^{-i c} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )}{2 e}+\frac{b^2 e^{2 i c} 2^{-\frac{m}{2}-\frac{7}{2}} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )}{e}+\frac{b^2 e^{-2 i c} 2^{-\frac{m}{2}-\frac{7}{2}} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )}{e}+\frac{\left (2 a^2+b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.2631, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{i a b e^{i c} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )}{2 e}-\frac{i a b e^{-i c} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )}{2 e}+\frac{b^2 e^{2 i c} 2^{-\frac{m}{2}-\frac{7}{2}} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )}{e}+\frac{b^2 e^{-2 i c} 2^{-\frac{m}{2}-\frac{7}{2}} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )}{e}+\frac{\left (2 a^2+b^2\right ) (e x)^{m+1}}{2 e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/2)*a*b*E^(I*c)*(e*x)^(1 + m)*((-I)*d*x^2)^((-1 - m)/2)*Gamma
[(1 + m)/2, (-I)*d*x^2])/e - ((I/2)*a*b*(e*x)^(1 + m)*(I*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, I*d*x^2])/(e*E^(
I*c)) + (2^(-7/2 - m/2)*b^2*E^((2*I)*c)*(e*x)^(1 + m)*((-I)*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (-2*I)*d*x^2]
)/e + (2^(-7/2 - m/2)*b^2*(e*x)^(1 + m)*(I*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (2*I)*d*x^2])/(e*E^((2*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^2\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^2\right )+2 a b (e x)^m \sin \left (c+d x^2\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^2\right )+2 a b (e x)^m \sin \left (c+d x^2\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^2\right ) \, dx-\frac{1}{2} b^2 \int (e x)^m \cos \left (2 c+2 d x^2\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^2} (e x)^m \, dx-(i a b) \int e^{i c+i d x^2} (e x)^m \, dx-\frac{1}{4} b^2 \int e^{-2 i c-2 i d x^2} (e x)^m \, dx-\frac{1}{4} b^2 \int e^{2 i c+2 i d x^2} (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^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},-i d x^2\right )}{2 e}-\frac{i a b e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},i d x^2\right )}{2 e}+\frac{2^{-\frac{7}{2}-\frac{m}{2}} b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},-2 i d x^2\right )}{e}+\frac{2^{-\frac{7}{2}-\frac{m}{2}} b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac{1}{2} (-1-m)} \Gamma \left (\frac{1+m}{2},2 i d x^2\right )}{e}\\ \end{align*}

Mathematica [A]  time = 6.51348, size = 551, normalized size = 1.97 \[ \frac{2^{\frac{1}{2} (-m-7)} x \left (d^2 x^4\right )^{\frac{1}{2} (-m-1)} (e x)^m \left (-i a b 2^{\frac{m+5}{2}} (m+1) (\cos (c)-i \sin (c)) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )+i a b 2^{\frac{m+5}{2}} (m+1) (\cos (c)+i \sin (c)) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )+b^2 \cos (2 c) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )+b^2 m \cos (2 c) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )+b^2 \cos (2 c) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )+b^2 m \cos (2 c) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )-i b^2 \sin (2 c) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )-i b^2 m \sin (2 c) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},2 i d x^2\right )+i b^2 \sin (2 c) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )+i b^2 m \sin (2 c) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-2 i d x^2\right )+a^2 2^{\frac{m+7}{2}} \left (d^2 x^4\right )^{\frac{m+1}{2}}+b^2 2^{\frac{m+5}{2}} \left (d^2 x^4\right )^{\frac{m+1}{2}}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^((-7 - m)/2)*x*(e*x)^m*(d^2*x^4)^((-1 - m)/2)*(2^((7 + m)/2)*a^2*(d^2*x^4)^((1 + m)/2) + 2^((5 + m)/2)*b^2*
(d^2*x^4)^((1 + m)/2) + b^2*(I*d*x^2)^((1 + m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (-2*I)*d*x^2] + b^2*m*(I*d*x^2)^((
1 + m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (-2*I)*d*x^2] + b^2*((-I)*d*x^2)^((1 + m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (2*
I)*d*x^2] + b^2*m*((-I)*d*x^2)^((1 + m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (2*I)*d*x^2] - I*2^((5 + m)/2)*a*b*(1 + m
)*((-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, I*d*x^2]*(Cos[c] - I*Sin[c]) + I*2^((5 + m)/2)*a*b*(1 + m)*(I*d*x^
2)^((1 + m)/2)*Gamma[(1 + m)/2, (-I)*d*x^2]*(Cos[c] + I*Sin[c]) + I*b^2*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2,
(-2*I)*d*x^2]*Sin[2*c] + I*b^2*m*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-2*I)*d*x^2]*Sin[2*c] - I*b^2*((-I)*
d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (2*I)*d*x^2]*Sin[2*c] - I*b^2*m*((-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2,
(2*I)*d*x^2]*Sin[2*c]))/(1 + m)

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Maple [F]  time = 0.374, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( a+b\sin \left ( d{x}^{2}+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03255, size = 555, normalized size = 1.99 \begin{align*} \frac{8 \,{\left (2 \, a^{2} + b^{2}\right )} \left (e x\right )^{m} d x +{\left (-i \, b^{2} e m - i \, b^{2} e\right )} e^{\left (-\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{2 i \, d}{e^{2}}\right ) - 2 i \, c\right )} \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, 2 i \, d x^{2}\right ) - 8 \,{\left (a b e m + a b e\right )} e^{\left (-\frac{1}{2} \,{\left (m - 1\right )} \log \left (\frac{i \, d}{e^{2}}\right ) - i \, c\right )} \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, i \, d x^{2}\right ) - 8 \,{\left (a b e m + a b e\right )} e^{\left (-\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{i \, d}{e^{2}}\right ) + i \, c\right )} \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -i \, d x^{2}\right ) +{\left (i \, b^{2} e m + i \, b^{2} e\right )} e^{\left (-\frac{1}{2} \,{\left (m - 1\right )} \log \left (-\frac{2 i \, d}{e^{2}}\right ) + 2 i \, c\right )} \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -2 i \, d x^{2}\right )}{16 \,{\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^2+c))^2,x, algorithm="fricas")

[Out]

1/16*(8*(2*a^2 + b^2)*(e*x)^m*d*x + (-I*b^2*e*m - I*b^2*e)*e^(-1/2*(m - 1)*log(2*I*d/e^2) - 2*I*c)*gamma(1/2*m
+ 1/2, 2*I*d*x^2) - 8*(a*b*e*m + a*b*e)*e^(-1/2*(m - 1)*log(I*d/e^2) - I*c)*gamma(1/2*m + 1/2, I*d*x^2) - 8*(
a*b*e*m + a*b*e)*e^(-1/2*(m - 1)*log(-I*d/e^2) + I*c)*gamma(1/2*m + 1/2, -I*d*x^2) + (I*b^2*e*m + I*b^2*e)*e^(
-1/2*(m - 1)*log(-2*I*d/e^2) + 2*I*c)*gamma(1/2*m + 1/2, -2*I*d*x^2))/(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^{2} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x^{2} + c\right ) + a\right )}^{2} \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^2+c))^2,x, algorithm="giac")

[Out]

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