\(\int \frac {e^{a+b x+c x^2} (b+2 c x)}{(a+b x+c x^2)^{5/2}} \, dx\) [633]

Optimal result

Integrand size = 33, antiderivative size = 85 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {4 e^{a+b x+c x^2}}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6839, 2208, 2211, 2235} \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {4}{3} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right )-\frac {4 e^{a+b x+c x^2}}{3 \sqrt {a+b x+c x^2}}-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[In]

Int[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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]

Rule 6839

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]

Rubi steps \begin{align*} \text {integral}& = \text {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}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac {2}{3} \text {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}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {4 e^{a+b x+c x^2}}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} \text {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}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {4 e^{a+b x+c x^2}}{3 \sqrt {a+b x+c x^2}}+\frac {8}{3} \text {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}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {4 e^{a+b x+c x^2}}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 3.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \left (e^{a+x (b+c x)} (1+2 (a+x (b+c x)))+2 (-a-x (b+c x))^{3/2} \Gamma \left (\frac {1}{2},-a-x (b+c x)\right )\right )}{3 (a+x (b+c x))^{3/2}} \]

[In]

Integrate[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {4 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{3}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{3 \sqrt {c \,x^{2}+b x +a}}\)	\(70\)
default	\(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {4 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{3}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{3 \sqrt {c \,x^{2}+b x +a}}\)	\(70\)

[In]

int(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 17.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {\left (\frac {4 \sqrt {\pi } \operatorname {erfc}{\left (\sqrt {- a - b x - c x^{2}} \right )}}{3} - \frac {\left (- \frac {4 a}{3} - \frac {4 b x}{3} - \frac {4 c x^{2}}{3} - \frac {2}{3}\right ) e^{a + b x + c x^{2}}}{\left (- a - b x - c x^{2}\right )^{\frac {3}{2}}}\right ) \left (- a - b x - c x^{2}\right )^{\frac {5}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}} \]

[In]

integrate(exp(c*x**2+b*x+a)*(2*c*x+b)/(c*x**2+b*x+a)**(5/2),x)

[Out]

(4*sqrt(pi)*erfc(sqrt(-a - b*x - c*x**2))/3 - (-4*a/3 - 4*b*x/3 - 4*c*x**2/3 - 2/3)*exp(a + b*x + c*x**2)/(-a
- b*x - c*x**2)**(3/2))*(-a - b*x - c*x**2)**(5/2)/(a + b*x + c*x**2)**(5/2)

Maxima [F]

\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*x + b)*e^(c*x^2 + b*x + a)/(c*x^2 + b*x + a)^(5/2), x)

Giac [F]

\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*x + b)*e^(c*x^2 + b*x + a)/(c*x^2 + b*x + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.22 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {{\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (2\,c\,x^2+2\,b\,x+2\,a\right )+4\,{\mathrm {e}}^{c\,x^2+b\,x+a}\,{\left (c\,x^2+b\,x+a\right )}^2-4\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )\,{\left (-c\,x^2-b\,x-a\right )}^{5/2}}{3\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \]

[In]

int((exp(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^(5/2),x)

[Out]

-(exp(a + b*x + c*x^2)*(2*a + 2*b*x + 2*c*x^2) + 4*exp(a + b*x + c*x^2)*(a + b*x + c*x^2)^2 - 4*pi^(1/2)*erfc(
(- a - b*x - c*x^2)^(1/2))*(- a - b*x - c*x^2)^(5/2))/(3*(a + b*x + c*x^2)^(5/2))