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

Optimal. Leaf size=115 \[ \frac{8}{15} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )-\frac{8 e^{a+b x+c x^2}}{15 \sqrt{a+b x+c x^2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

(-2*E^(a + b*x + c*x^2))/(5*(a + b*x + c*x^2)^(5/2)) - (4*E^(a + b*x + c*x^2))/(15*(a + b*x + c*x^2)^(3/2)) -
(8*E^(a + b*x + c*x^2))/(15*Sqrt[a + b*x + c*x^2]) + (8*Sqrt[Pi]*Erfi[Sqrt[a + b*x + c*x^2]])/15

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Rubi [A]  time = 0.35466, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {6707, 2177, 2180, 2204} \[ \frac{8}{15} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )-\frac{8 e^{a+b x+c x^2}}{15 \sqrt{a+b x+c x^2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(a + b*x + c*x^2))/(5*(a + b*x + c*x^2)^(5/2)) - (4*E^(a + b*x + c*x^2))/(15*(a + b*x + c*x^2)^(3/2)) -
(8*E^(a + b*x + c*x^2))/(15*Sqrt[a + b*x + c*x^2]) + (8*Sqrt[Pi]*Erfi[Sqrt[a + b*x + c*x^2]])/15

Rule 6707

Int[(F_)^(v_)*(u_)*(w_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Dist[q, Subst[Int[x^m*F^x,
x], x, v], x] /;  !FalseQ[q]] /; FreeQ[{F, m}, x] && EqQ[w, v]

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{e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx &=\operatorname{Subst}\left (\int \frac{e^x}{x^{7/2}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}+\frac{2}{5} \operatorname{Subst}\left (\int \frac{e^x}{x^{5/2}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}+\frac{4}{15} \operatorname{Subst}\left (\int \frac{e^x}{x^{3/2}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 e^{a+b x+c x^2}}{15 \sqrt{a+b x+c x^2}}+\frac{8}{15} \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{x}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 e^{a+b x+c x^2}}{15 \sqrt{a+b x+c x^2}}+\frac{16}{15} \operatorname{Subst}\left (\int e^{x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}-\frac{4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 e^{a+b x+c x^2}}{15 \sqrt{a+b x+c x^2}}+\frac{8}{15} \sqrt{\pi } \text{erfi}\left (\sqrt{a+b x+c x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.151417, size = 91, normalized size = 0.79 \[ \frac{8 (-a-x (b+c x))^{5/2} \text{Gamma}\left (\frac{1}{2},-a-x (b+c x)\right )-2 e^{a+x (b+c x)} \left (4 (a+x (b+c x))^2+2 (a+x (b+c x))+3\right )}{15 (a+x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(a + x*(b + c*x))*(3 + 2*(a + x*(b + c*x)) + 4*(a + x*(b + c*x))^2) + 8*(-a - x*(b + c*x))^(5/2)*Gamma[1
/2, -a - x*(b + c*x)])/(15*(a + x*(b + c*x))^(5/2))

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Maple [A]  time = 0.051, size = 95, normalized size = 0.8 \begin{align*} -{\frac{2\,{{\rm e}^{c{x}^{2}+bx+a}}}{5} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{5}{2}}}}-{\frac{4\,{{\rm e}^{c{x}^{2}+bx+a}}}{15} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,\sqrt{\pi }}{15}{\it erfi} \left ( \sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{8\,{{\rm e}^{c{x}^{2}+bx+a}}}{15}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^(7/2),x)

[Out]

-2/5*exp(c*x^2+b*x+a)/(c*x^2+b*x+a)^(5/2)-4/15*exp(c*x^2+b*x+a)/(c*x^2+b*x+a)^(3/2)+8/15*erfi((c*x^2+b*x+a)^(1
/2))*Pi^(1/2)-8/15*exp(c*x^2+b*x+a)/(c*x^2+b*x+a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{c^{4} x^{8} + 4 \, b c^{3} x^{7} + 2 \,{\left (3 \, b^{2} c^{2} + 2 \, a c^{3}\right )} x^{6} + 4 \,{\left (b^{3} c + 3 \, a b c^{2}\right )} x^{5} + 4 \, a^{3} b x +{\left (b^{4} + 12 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} x^{4} + a^{4} + 4 \,{\left (a b^{3} + 3 \, a^{2} b c\right )} x^{3} + 2 \,{\left (3 \, a^{2} b^{2} + 2 \, a^{3} c\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*e^(c*x^2 + b*x + a)/(c^4*x^8 + 4*b*c^3*x^7 + 2*(3*b^2*c^2 + 2*a*c^3
)*x^6 + 4*(b^3*c + 3*a*b*c^2)*x^5 + 4*a^3*b*x + (b^4 + 12*a*b^2*c + 6*a^2*c^2)*x^4 + a^4 + 4*(a*b^3 + 3*a^2*b*
c)*x^3 + 2*(3*a^2*b^2 + 2*a^3*c)*x^2), x)

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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(exp(c*x**2+b*x+a)*(2*c*x+b)/(c*x**2+b*x+a)**(7/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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