3.47 \(\int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac {16 \sqrt {\pi } b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F) F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{105 e^{9/2}}-\frac {16 b^3 c^3 \log ^3(F) F^{c (a+b x)}}{105 e^4 \sqrt {d+e x}}-\frac {8 b^2 c^2 \log ^2(F) F^{c (a+b x)}}{105 e^3 (d+e x)^{3/2}}-\frac {4 b c \log (F) F^{c (a+b x)}}{35 e^2 (d+e x)^{5/2}}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}} \]

[Out]

-2/7*F^(c*(b*x+a))/e/(e*x+d)^(7/2)-4/35*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d)^(5/2)-8/105*b^2*c^2*F^(c*(b*x+a))*
ln(F)^2/e^3/(e*x+d)^(3/2)+16/105*b^(7/2)*c^(7/2)*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)^(7/2)*Pi^(1/2)/e^(9/2)-16/105*b^3*c^3*F^(c*(b*x+a))*ln(F)^3/e^4/(e*x+d)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2177, 2180, 2204} \[ \frac {16 \sqrt {\pi } b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F) F^{c \left (a-\frac {b d}{e}\right )} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{105 e^{9/2}}-\frac {16 b^3 c^3 \log ^3(F) F^{c (a+b x)}}{105 e^4 \sqrt {d+e x}}-\frac {8 b^2 c^2 \log ^2(F) F^{c (a+b x)}}{105 e^3 (d+e x)^{3/2}}-\frac {4 b c \log (F) F^{c (a+b x)}}{35 e^2 (d+e x)^{5/2}}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d + e*x)^(9/2),x]

[Out]

(-2*F^(c*(a + b*x)))/(7*e*(d + e*x)^(7/2)) - (4*b*c*F^(c*(a + b*x))*Log[F])/(35*e^2*(d + e*x)^(5/2)) - (8*b^2*
c^2*F^(c*(a + b*x))*Log[F]^2)/(105*e^3*(d + e*x)^(3/2)) - (16*b^3*c^3*F^(c*(a + b*x))*Log[F]^3)/(105*e^4*Sqrt[
d + e*x]) + (16*b^(7/2)*c^(7/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])
/Sqrt[e]]*Log[F]^(7/2))/(105*e^(9/2))

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)^{9/2}} \, dx &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}+\frac {(2 b c \log (F)) \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx}{7 e}\\ &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}+\frac {\left (4 b^2 c^2 \log ^2(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx}{35 e^2}\\ &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}+\frac {\left (8 b^3 c^3 \log ^3(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^{3/2}} \, dx}{105 e^3}\\ &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {\left (16 b^4 c^4 \log ^4(F)\right ) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}} \, dx}{105 e^4}\\ &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {\left (32 b^4 c^4 \log ^4(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 )}{105 e^5}\\ &=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {16 b^{7/2} c^{7/2} 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 {7}{2}}(F)}{105 e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 144, normalized size = 0.72 \[ \frac {2 \left (2 b c \log (F) (d+e x) \left (-2 b c \log (F) (d+e x) \left (F^{c (a+b x)} (2 b c \log (F) (d+e x)+e)+2 e F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right )\right )-3 e^2 F^{c (a+b x)}\right )-15 e^3 F^{c (a+b x)}\right )}{105 e^4 (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d + e*x)^(9/2),x]

[Out]

(2*(-15*e^3*F^(c*(a + b*x)) + 2*b*c*(d + e*x)*Log[F]*(-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])))))/(105*e^4*(d + e*x)^(7/2))

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fricas [A]  time = 0.44, size = 321, normalized size = 1.60 \[ -\frac {2 \, {\left (\frac {8 \, \sqrt {\pi } {\left (b^{3} c^{3} e^{4} x^{4} + 4 \, b^{3} c^{3} d e^{3} x^{3} + 6 \, b^{3} c^{3} d^{2} e^{2} x^{2} + 4 \, b^{3} c^{3} d^{3} e x + b^{3} c^{3} d^{4}\right )} \sqrt {-\frac {b c \log \relax (F)}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \relax (F)}{e}}\right ) \log \relax (F)^{3}}{F^{\frac {b c d - a c e}{e}}} + {\left (8 \, {\left (b^{3} c^{3} e^{3} x^{3} + 3 \, b^{3} c^{3} d e^{2} x^{2} + 3 \, b^{3} c^{3} d^{2} e x + b^{3} c^{3} d^{3}\right )} \log \relax (F)^{3} + 15 \, e^{3} + 4 \, {\left (b^{2} c^{2} e^{3} x^{2} + 2 \, b^{2} c^{2} d e^{2} x + b^{2} c^{2} d^{2} e\right )} \log \relax (F)^{2} + 6 \, {\left (b c e^{3} x + b c d e^{2}\right )} \log \relax (F)\right )} \sqrt {e x + d} F^{b c x + a c}\right )}}{105 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(8*sqrt(pi)*(b^3*c^3*e^4*x^4 + 4*b^3*c^3*d*e^3*x^3 + 6*b^3*c^3*d^2*e^2*x^2 + 4*b^3*c^3*d^3*e*x + b^3*c^
3*d^4)*sqrt(-b*c*log(F)/e)*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))*log(F)^3/F^((b*c*d - a*c*e)/e) + (8*(b^3*c^3
*e^3*x^3 + 3*b^3*c^3*d*e^2*x^2 + 3*b^3*c^3*d^2*e*x + b^3*c^3*d^3)*log(F)^3 + 15*e^3 + 4*(b^2*c^2*e^3*x^2 + 2*b
^2*c^2*d*e^2*x + b^2*c^2*d^2*e)*log(F)^2 + 6*(b*c*e^3*x + b*c*d*e^2)*log(F))*sqrt(e*x + d)*F^(b*c*x + a*c))/(e
^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4*d^3*e^5*x + d^4*e^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(9/2),x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^(9/2), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {F^{\left (b x +a \right ) c}}{\left (e x +d \right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^((b*x+a)*c)/(e*x+d)^(9/2),x)

[Out]

int(F^((b*x+a)*c)/(e*x+d)^(9/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^(9/2),x, algorithm="maxima")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(d + e*x)^(9/2),x)

[Out]

int(F^(c*(a + b*x))/(d + e*x)^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(e*x+d)**(9/2),x)

[Out]

Timed out

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