3.40 \(\int F^{c (a+b x)} (d+e x)^{5/2} \, dx\)

Optimal. Leaf size=173 \[ -\frac{15 \sqrt{\pi } e^{5/2} F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac{7}{2}}(F)}+\frac{15 e^2 \sqrt{d+e x} F^{c (a+b x)}}{4 b^3 c^3 \log ^3(F)}-\frac{5 e (d+e x)^{3/2} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac{(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)} \]

[Out]

(-15*e^(5/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(8*b^(7/
2)*c^(7/2)*Log[F]^(7/2)) + (15*e^2*F^(c*(a + b*x))*Sqrt[d + e*x])/(4*b^3*c^3*Log[F]^3) - (5*e*F^(c*(a + b*x))*
(d + e*x)^(3/2))/(2*b^2*c^2*Log[F]^2) + (F^(c*(a + b*x))*(d + e*x)^(5/2))/(b*c*Log[F])

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Rubi [A]  time = 0.165517, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2176, 2180, 2204} \[ -\frac{15 \sqrt{\pi } e^{5/2} F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac{7}{2}}(F)}+\frac{15 e^2 \sqrt{d+e x} F^{c (a+b x)}}{4 b^3 c^3 \log ^3(F)}-\frac{5 e (d+e x)^{3/2} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac{(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*e^(5/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(8*b^(7/
2)*c^(7/2)*Log[F]^(7/2)) + (15*e^2*F^(c*(a + b*x))*Sqrt[d + e*x])/(4*b^3*c^3*Log[F]^3) - (5*e*F^(c*(a + b*x))*
(d + e*x)^(3/2))/(2*b^2*c^2*Log[F]^2) + (F^(c*(a + b*x))*(d + e*x)^(5/2))/(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)^{5/2} \, dx &=\frac{F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac{(5 e) \int F^{c (a+b x)} (d+e x)^{3/2} \, dx}{2 b c \log (F)}\\ &=-\frac{5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}+\frac{\left (15 e^2\right ) \int F^{c (a+b x)} \sqrt{d+e x} \, dx}{4 b^2 c^2 \log ^2(F)}\\ &=\frac{15 e^2 F^{c (a+b x)} \sqrt{d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac{5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac{\left (15 e^3\right ) \int \frac{F^{c (a+b x)}}{\sqrt{d+e x}} \, dx}{8 b^3 c^3 \log ^3(F)}\\ &=\frac{15 e^2 F^{c (a+b x)} \sqrt{d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac{5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac{\left (15 e^2\right ) \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 )}{4 b^3 c^3 \log ^3(F)}\\ &=-\frac{15 e^{5/2} 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 )}{8 b^{7/2} c^{7/2} \log ^{\frac{7}{2}}(F)}+\frac{15 e^2 F^{c (a+b x)} \sqrt{d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac{5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac{F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0266816, size = 72, normalized size = 0.42 \[ \frac{e^2 \sqrt{d+e x} F^{c \left (a-\frac{b d}{e}\right )} \text{Gamma}\left (\frac{7}{2},-\frac{b c \log (F) (d+e x)}{e}\right )}{b^3 c^3 \log ^3(F) \sqrt{-\frac{b c \log (F) (d+e x)}{e}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*F^(c*(a - (b*d)/e))*Sqrt[d + e*x]*Gamma[7/2, -((b*c*(d + e*x)*Log[F])/e)])/(b^3*c^3*Log[F]^3*Sqrt[-((b*c*
(d + e*x)*Log[F])/e)])

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*(e*x+d)^(5/2),x)

[Out]

int(F^(c*(b*x+a))*(e*x+d)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)*F^((b*x + a)*c), x)

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Fricas [A]  time = 1.54212, size = 383, normalized size = 2.21 \begin{align*} \frac{\frac{15 \, \sqrt{\pi } \sqrt{-\frac{b c \log \left (F\right )}{e}} e^{3} \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 (15 \, b c e^{2} \log \left (F\right ) + 4 \,{\left (b^{3} c^{3} e^{2} x^{2} + 2 \, b^{3} c^{3} d e x + b^{3} c^{3} d^{2}\right )} \log \left (F\right )^{3} - 10 \,{\left (b^{2} c^{2} e^{2} x + b^{2} c^{2} d e\right )} \log \left (F\right )^{2}\right )} \sqrt{e x + d} F^{b c x + a c}}{8 \, b^{4} c^{4} \log \left (F\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

1/8*(15*sqrt(pi)*sqrt(-b*c*log(F)/e)*e^3*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))/F^((b*c*d - a*c*e)/e) + 2*(15*
b*c*e^2*log(F) + 4*(b^3*c^3*e^2*x^2 + 2*b^3*c^3*d*e*x + b^3*c^3*d^2)*log(F)^3 - 10*(b^2*c^2*e^2*x + b^2*c^2*d*
e)*log(F)^2)*sqrt(e*x + d)*F^(b*c*x + a*c))/(b^4*c^4*log(F)^4)

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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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.33364, size = 779, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(5/2),x, algorithm="giac")

[Out]

1/8*(4*d^2*(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))) - 4*d*(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)) + (sqrt(pi)*(4*b^2*c^2*d^2*e*log(F)^2 + 12*b*c*d*e^2*log(F) +
15*e^3)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*e^(-(b*c*d*log(F) - a*c*e*log(F) + 2*e)*e^(-1) + 1)/(sq
rt(-b*c*e*log(F))*b^3*c^3*log(F)^3) + 2*(4*(x*e + d)^(5/2)*b^2*c^2*e*log(F)^2 - 8*(x*e + d)^(3/2)*b^2*c^2*d*e*
log(F)^2 + 4*sqrt(x*e + d)*b^2*c^2*d^2*e*log(F)^2 - 10*(x*e + d)^(3/2)*b*c*e^2*log(F) + 12*sqrt(x*e + d)*b*c*d
*e^2*log(F) + 15*sqrt(x*e + d)*e^3)*e^(((x*e + d)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F) - 2*e)*e^(-1))/(b^3
*c^3*log(F)^3))*e^2)*e^(-1)