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

Optimal. Leaf size=134 \[ \frac{i b e^{i c} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )}{4 e}-\frac{i b e^{-i c} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )}{4 e}+\frac{a (e x)^{m+1}}{e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.118672, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {14, 3389, 2218} \[ \frac{i b e^{i c} \left (-i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )}{4 e}-\frac{i b e^{-i c} \left (i d x^2\right )^{\frac{1}{2} (-m-1)} (e x)^{m+1} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )}{4 e}+\frac{a (e x)^{m+1}}{e (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*(a + b*Sin[c + d*x^2]),x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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]

Rubi steps

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

Mathematica [A]  time = 1.57058, size = 149, normalized size = 1.11 \[ \frac{x \left (d^2 x^4\right )^{\frac{1}{2} (-m-1)} (e x)^m \left (-i b (m+1) (\cos (c)-i \sin (c)) \left (-i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},i d x^2\right )+i b (m+1) (\cos (c)+i \sin (c)) \left (i d x^2\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{m+1}{2},-i d x^2\right )+4 a \left (d^2 x^4\right )^{\frac{m+1}{2}}\right )}{4 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.125, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( a+b\sin \left ( d{x}^{2}+c \right ) \right ) \, 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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69436, size = 263, normalized size = 1.96 \begin{align*} \frac{4 \, \left (e x\right )^{m} a d x -{\left (b e m + 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 (b e m + 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 )}{4 \,{\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)),x, algorithm="fricas")

[Out]

1/4*(4*(e*x)^m*a*d*x - (b*e*m + b*e)*e^(-1/2*(m - 1)*log(I*d/e^2) - I*c)*gamma(1/2*m + 1/2, I*d*x^2) - (b*e*m
+ b*e)*e^(-1/2*(m - 1)*log(-I*d/e^2) + I*c)*gamma(1/2*m + 1/2, -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 )\, 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)),x)

[Out]

Integral((e*x)**m*(a + b*sin(c + d*x**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 )} \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)),x, algorithm="giac")

[Out]

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