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

Optimal. Leaf size=97 \[ \frac {2 \sqrt {\pi } \sqrt {b} \sqrt {c} \sqrt {\log (F)} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}} \]

[Out]

-2*F^(c*(b*x+a))/e/(e*x+d)^(1/2)+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))*b^(
1/2)*c^(1/2)*Pi^(1/2)*ln(F)^(1/2)/e^(3/2)

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Rubi [A]  time = 0.09, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2177, 2180, 2204} \[ \frac {2 \sqrt {\pi } \sqrt {b} \sqrt {c} \sqrt {\log (F)} F^{c \left (a-\frac {b d}{e}\right )} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)^{3/2}} \, dx &=-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}+\frac {(2 b c \log (F)) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}} \, dx}{e}\\ &=-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}+\frac {(4 b c \log (F)) \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 )}{e^2}\\ &=-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}+\frac {2 \sqrt {b} \sqrt {c} 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 ) \sqrt {\log (F)}}{e^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 75, normalized size = 0.77 \[ -\frac {2 \left (F^{c (a+b x)}-F^{c \left (a-\frac {b d}{e}\right )} \sqrt {-\frac {b c \log (F) (d+e x)}{e}} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right )\right )}{e \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 90, normalized size = 0.93 \[ -\frac {2 \, {\left (\frac {\sqrt {\pi } {\left (e x + d\right )} \sqrt {-\frac {b c \log \relax (F)}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \relax (F)}{e}}\right )}{F^{\frac {b c d - a c e}{e}}} + \sqrt {e x + d} F^{b c x + a c}\right )}}{e^{2} x + d e} \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {F^{\left (b x +a \right ) c}}{\left (e x +d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

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(F^((b*x + a)*c)/(e*x + d)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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