3.41 \(\int F^{c (a+b x)} (d+e x)^{3/2} \, dx\)

Optimal. Leaf size=138 \[ \frac{3 \sqrt{\pi } e^{3/2} F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{4 b^{5/2} c^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 e \sqrt{d+e x} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac{(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)} \]

[Out]

(3*e^(3/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(4*b^(5/2)
*c^(5/2)*Log[F]^(5/2)) - (3*e*F^(c*(a + b*x))*Sqrt[d + e*x])/(2*b^2*c^2*Log[F]^2) + (F^(c*(a + b*x))*(d + e*x)
^(3/2))/(b*c*Log[F])

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Rubi [A]  time = 0.11937, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2176, 2180, 2204} \[ \frac{3 \sqrt{\pi } e^{3/2} F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{4 b^{5/2} c^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 e \sqrt{d+e x} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac{(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*e^(3/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(4*b^(5/2)
*c^(5/2)*Log[F]^(5/2)) - (3*e*F^(c*(a + b*x))*Sqrt[d + e*x])/(2*b^2*c^2*Log[F]^2) + (F^(c*(a + b*x))*(d + e*x)
^(3/2))/(b*c*Log[F])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rubi steps

\begin{align*} \int F^{c (a+b x)} (d+e x)^{3/2} \, dx &=\frac{F^{c (a+b x)} (d+e x)^{3/2}}{b c \log (F)}-\frac{(3 e) \int F^{c (a+b x)} \sqrt{d+e x} \, dx}{2 b c \log (F)}\\ &=-\frac{3 e F^{c (a+b x)} \sqrt{d+e x}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{3/2}}{b c \log (F)}+\frac{\left (3 e^2\right ) \int \frac{F^{c (a+b x)}}{\sqrt{d+e x}} \, dx}{4 b^2 c^2 \log ^2(F)}\\ &=-\frac{3 e F^{c (a+b x)} \sqrt{d+e x}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{3/2}}{b c \log (F)}+\frac{(3 e) \operatorname{Subst}\left (\int F^{c \left (a-\frac{b d}{e}\right )+\frac{b c x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{2 b^2 c^2 \log ^2(F)}\\ &=\frac{3 e^{3/2} F^{c \left (a-\frac{b d}{e}\right )} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{d+e x} \sqrt{\log (F)}}{\sqrt{e}}\right )}{4 b^{5/2} c^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 e F^{c (a+b x)} \sqrt{d+e x}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{3/2}}{b c \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0700684, size = 63, normalized size = 0.46 \[ -\frac{(d+e x)^{5/2} F^{c \left (a-\frac{b d}{e}\right )} \text{Gamma}\left (\frac{5}{2},-\frac{b c \log (F) (d+e x)}{e}\right )}{e \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((F^(c*(a - (b*d)/e))*(d + e*x)^(5/2)*Gamma[5/2, -((b*c*(d + e*x)*Log[F])/e)])/(e*(-((b*c*(d + e*x)*Log[F])/e
))^(5/2)))

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*(e*x+d)^(3/2),x)

[Out]

int(F^(c*(b*x+a))*(e*x+d)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{\frac{3}{2}} F^{{\left (b x + a\right )} c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(3/2)*F^((b*x + a)*c), x)

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Fricas [A]  time = 1.91564, size = 288, normalized size = 2.09 \begin{align*} -\frac{\frac{3 \, \sqrt{\pi } \sqrt{-\frac{b c \log \left (F\right )}{e}} e^{2} \operatorname{erf}\left (\sqrt{e x + d} \sqrt{-\frac{b c \log \left (F\right )}{e}}\right )}{F^{\frac{b c d - a c e}{e}}} + 2 \,{\left (3 \, b c e \log \left (F\right ) - 2 \,{\left (b^{2} c^{2} e x + b^{2} c^{2} d\right )} \log \left (F\right )^{2}\right )} \sqrt{e x + d} F^{b c x + a c}}{4 \, b^{3} c^{3} \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*sqrt(pi)*sqrt(-b*c*log(F)/e)*e^2*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))/F^((b*c*d - a*c*e)/e) + 2*(3*b
*c*e*log(F) - 2*(b^2*c^2*e*x + b^2*c^2*d)*log(F)^2)*sqrt(e*x + d)*F^(b*c*x + a*c))/(b^3*c^3*log(F)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*(e*x+d)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.19088, size = 408, normalized size = 2.96 \begin{align*} \frac{1}{4} \,{\left (2 \, d{\left (\frac{\sqrt{\pi } \operatorname{erf}\left (-\sqrt{-b c e \log \left (F\right )} \sqrt{x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \left (F\right ) - a c e \log \left (F\right )\right )} e^{\left (-1\right )} + 2\right )}}{\sqrt{-b c e \log \left (F\right )} b c \log \left (F\right )} + \frac{2 \, \sqrt{x e + d} e^{\left ({\left ({\left (x e + d\right )} b c \log \left (F\right ) - b c d \log \left (F\right ) + a c e \log \left (F\right )\right )} e^{\left (-1\right )} + 1\right )}}{b c \log \left (F\right )}\right )} - \frac{\sqrt{\pi }{\left (2 \, b c d e \log \left (F\right ) + 3 \, e^{2}\right )} \operatorname{erf}\left (-\sqrt{-b c e \log \left (F\right )} \sqrt{x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \left (F\right ) - a c e \log \left (F\right )\right )} e^{\left (-1\right )} + 1\right )}}{\sqrt{-b c e \log \left (F\right )} b^{2} c^{2} \log \left (F\right )^{2}} + \frac{2 \,{\left (2 \,{\left (x e + d\right )}^{\frac{3}{2}} b c e \log \left (F\right ) - 2 \, \sqrt{x e + d} b c d e \log \left (F\right ) - 3 \, \sqrt{x e + d} e^{2}\right )} e^{\left ({\left ({\left (x e + d\right )} b c \log \left (F\right ) - b c d \log \left (F\right ) + a c e \log \left (F\right )\right )} e^{\left (-1\right )}\right )}}{b^{2} c^{2} \log \left (F\right )^{2}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(3/2),x, algorithm="giac")

[Out]

1/4*(2*d*(sqrt(pi)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*e^(-(b*c*d*log(F) - a*c*e*log(F))*e^(-1) + 2
)/(sqrt(-b*c*e*log(F))*b*c*log(F)) + 2*sqrt(x*e + d)*e^(((x*e + d)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F))*e
^(-1) + 1)/(b*c*log(F))) - sqrt(pi)*(2*b*c*d*e*log(F) + 3*e^2)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*
e^(-(b*c*d*log(F) - a*c*e*log(F))*e^(-1) + 1)/(sqrt(-b*c*e*log(F))*b^2*c^2*log(F)^2) + 2*(2*(x*e + d)^(3/2)*b*
c*e*log(F) - 2*sqrt(x*e + d)*b*c*d*e*log(F) - 3*sqrt(x*e + d)*e^2)*e^(((x*e + d)*b*c*log(F) - b*c*d*log(F) + a
*c*e*log(F))*e^(-1))/(b^2*c^2*log(F)^2))*e^(-1)