3.46 \(\int \frac{F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac{8 \sqrt{\pi } b^{5/2} c^{5/2} \log ^{\frac{5}{2}}(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{15 e^{7/2}}-\frac{8 b^2 c^2 \log ^2(F) F^{c (a+b x)}}{15 e^3 \sqrt{d+e x}}-\frac{4 b c \log (F) F^{c (a+b x)}}{15 e^2 (d+e x)^{3/2}}-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.164663, antiderivative size = 165, 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 = {2177, 2180, 2204} \[ \frac{8 \sqrt{\pi } b^{5/2} c^{5/2} \log ^{\frac{5}{2}}(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c} \sqrt{\log (F)} \sqrt{d+e x}}{\sqrt{e}}\right )}{15 e^{7/2}}-\frac{8 b^2 c^2 \log ^2(F) F^{c (a+b x)}}{15 e^3 \sqrt{d+e x}}-\frac{4 b c \log (F) F^{c (a+b x)}}{15 e^2 (d+e x)^{3/2}}-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d + e*x)^(7/2),x]

[Out]

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

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && 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 \frac{F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx &=-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}+\frac{(2 b c \log (F)) \int \frac{F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx}{5 e}\\ &=-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}-\frac{4 b c F^{c (a+b x)} \log (F)}{15 e^2 (d+e x)^{3/2}}+\frac{\left (4 b^2 c^2 \log ^2(F)\right ) \int \frac{F^{c (a+b x)}}{(d+e x)^{3/2}} \, dx}{15 e^2}\\ &=-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}-\frac{4 b c F^{c (a+b x)} \log (F)}{15 e^2 (d+e x)^{3/2}}-\frac{8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{15 e^3 \sqrt{d+e x}}+\frac{\left (8 b^3 c^3 \log ^3(F)\right ) \int \frac{F^{c (a+b x)}}{\sqrt{d+e x}} \, dx}{15 e^3}\\ &=-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}-\frac{4 b c F^{c (a+b x)} \log (F)}{15 e^2 (d+e x)^{3/2}}-\frac{8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{15 e^3 \sqrt{d+e x}}+\frac{\left (16 b^3 c^3 \log ^3(F)\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 )}{15 e^4}\\ &=-\frac{2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}-\frac{4 b c F^{c (a+b x)} \log (F)}{15 e^2 (d+e x)^{3/2}}-\frac{8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{15 e^3 \sqrt{d+e x}}+\frac{8 b^{5/2} c^{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 ) \log ^{\frac{5}{2}}(F)}{15 e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.137881, size = 118, normalized size = 0.72 \[ \frac{2 \left (-2 b c \log (F) (d+e x) \left (2 e F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{b c \log (F) (d+e x)}{e}\right )+F^{c (a+b x)} (2 b c \log (F) (d+e x)+e)\right )-3 e^2 F^{c (a+b x)}\right )}{15 e^3 (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d + e*x)^(7/2),x]

[Out]

(2*(-3*e^2*F^(c*(a + b*x)) - 2*b*c*(d + e*x)*Log[F]*(2*e*F^(c*(a - (b*d)/e))*Gamma[1/2, -((b*c*(d + e*x)*Log[F
])/e)]*(-((b*c*(d + e*x)*Log[F])/e))^(3/2) + F^(c*(a + b*x))*(e + 2*b*c*(d + e*x)*Log[F]))))/(15*e^3*(d + e*x)
^(5/2))

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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{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(e*x+d)^(7/2),x)

[Out]

int(F^(c*(b*x+a))/(e*x+d)^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^(7/2), x)

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Fricas [A]  time = 1.56885, size = 497, normalized size = 3.01 \begin{align*} -\frac{2 \,{\left (\frac{4 \, \sqrt{\pi }{\left (b^{2} c^{2} e^{3} x^{3} + 3 \, b^{2} c^{2} d e^{2} x^{2} + 3 \, b^{2} c^{2} d^{2} e x + b^{2} c^{2} d^{3}\right )} \sqrt{-\frac{b c \log \left (F\right )}{e}} \operatorname{erf}\left (\sqrt{e x + d} \sqrt{-\frac{b c \log \left (F\right )}{e}}\right ) \log \left (F\right )^{2}}{F^{\frac{b c d - a c e}{e}}} +{\left (4 \,{\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} \log \left (F\right )^{2} + 3 \, e^{2} + 2 \,{\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} \sqrt{e x + d} F^{b c x + a c}\right )}}{15 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(4*sqrt(pi)*(b^2*c^2*e^3*x^3 + 3*b^2*c^2*d*e^2*x^2 + 3*b^2*c^2*d^2*e*x + b^2*c^2*d^3)*sqrt(-b*c*log(F)/e
)*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))*log(F)^2/F^((b*c*d - a*c*e)/e) + (4*(b^2*c^2*e^2*x^2 + 2*b^2*c^2*d*e*
x + b^2*c^2*d^2)*log(F)^2 + 3*e^2 + 2*(b*c*e^2*x + b*c*d*e)*log(F))*sqrt(e*x + d)*F^(b*c*x + a*c))/(e^6*x^3 +
3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^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)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^(7/2), x)