3.1.46 \(\int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx\) [46]

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

[Out]

-2/5*F^(c*(b*x+a))/e/(e*x+d)^(5/2)-4/15*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d)^(3/2)+8/15*b^(5/2)*c^(5/2)*F^(c*(a
-b*d/e))*erfi(b^(1/2)*c^(1/2)*(e*x+d)^(1/2)*ln(F)^(1/2)/e^(1/2))*ln(F)^(5/2)*Pi^(1/2)/e^(7/2)-8/15*b^2*c^2*F^(
c*(b*x+a))*ln(F)^2/e^3/(e*x+d)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 165, normalized size of antiderivative = 1.00, 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 = {2208, 2211, 2235} \begin {gather*} \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}} \end {gather*}

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 ) \text {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*}

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Mathematica [A]
time = 0.27, size = 118, normalized size = 0.72 \begin {gather*} \frac {2 \left (-3 e^2 F^{c (a+b x)}-2 b c (d+e x) \log (F) \left (2 e F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{3/2}+F^{c (a+b x)} (e+2 b c (d+e x) \log (F))\right )\right )}{15 e^3 (d+e x)^{5/2}} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {F^{c \left (b x +a \right )}}{\left (e x +d \right )^{\frac {7}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)/(x*e + d)^(7/2), x)

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Fricas [A]
time = 0.38, size = 224, normalized size = 1.36 \begin {gather*} -\frac {2 \, {\left (\frac {4 \, \sqrt {\pi } {\left (b^{2} c^{2} x^{3} e^{3} + 3 \, b^{2} c^{2} d x^{2} e^{2} + 3 \, b^{2} c^{2} d^{2} x e + b^{2} c^{2} d^{3}\right )} \sqrt {-b c e^{\left (-1\right )} \log \left (F\right )} \operatorname {erf}\left (\sqrt {-b c e^{\left (-1\right )} \log \left (F\right )} \sqrt {x e + d}\right ) \log \left (F\right )^{2}}{F^{{\left (b c d - a c e\right )} e^{\left (-1\right )}}} + {\left (4 \, {\left (b^{2} c^{2} x^{2} e^{2} + 2 \, b^{2} c^{2} d x e + b^{2} c^{2} d^{2}\right )} \log \left (F\right )^{2} + 2 \, {\left (b c x e^{2} + b c d e\right )} \log \left (F\right ) + 3 \, e^{2}\right )} \sqrt {x e + d} F^{b c x + a c}\right )}}{15 \, {\left (x^{3} e^{6} + 3 \, d x^{2} e^{5} + 3 \, d^{2} x e^{4} + d^{3} e^{3}\right )}} \end {gather*}

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*x^3*e^3 + 3*b^2*c^2*d*x^2*e^2 + 3*b^2*c^2*d^2*x*e + b^2*c^2*d^3)*sqrt(-b*c*e^(-1)*l
og(F))*erf(sqrt(-b*c*e^(-1)*log(F))*sqrt(x*e + d))*log(F)^2/F^((b*c*d - a*c*e)*e^(-1)) + (4*(b^2*c^2*x^2*e^2 +
2*b^2*c^2*d*x*e + b^2*c^2*d^2)*log(F)^2 + 2*(b*c*x*e^2 + b*c*d*e)*log(F) + 3*e^2)*sqrt(x*e + d)*F^(b*c*x + a*
c))/(x^3*e^6 + 3*d*x^2*e^5 + 3*d^2*x*e^4 + d^3*e^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x*e + d)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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