3.7.34 \(\int \frac {e^{a+b x+c x^2} (b+2 c x)}{(a+b x+c x^2)^{7/2}} \, dx\) [634]

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

[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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Rubi [A]
time = 0.24, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 = {6839, 2208, 2211, 2235} \begin {gather*} \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}} \end {gather*}

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 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*} \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx &=\text {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} \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}}{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} \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}}{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} \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}}{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} \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}}{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*}

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Mathematica [A]
time = 4.91, size = 91, normalized size = 0.79 \begin {gather*} \frac {-2 e^{a+x (b+c x)} \left (3+2 (a+x (b+c x))+4 (a+x (b+c x))^2\right )+8 (-a-x (b+c x))^{5/2} \Gamma \left (\frac {1}{2},-a-x (b+c x)\right )}{15 (a+x (b+c x))^{5/2}} \end {gather*}

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 = 0.10, size = 95, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{5 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 \erfi \left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{15}-\frac {8 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \sqrt {c \,x^{2}+b x +a}}\)	\(95\)
default	\(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{5 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 \erfi \left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{15}-\frac {8 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \sqrt {c \,x^{2}+b x +a}}\)	\(95\)

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,method=_RETURNVERBOSE)

[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 = 0.00, 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(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 = 0.00, 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(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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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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(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)

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Mupad [B]
time = 4.66, size = 129, normalized size = 1.12 \begin {gather*} -\frac {{\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (6\,c\,x^2+6\,b\,x+6\,a\right )+4\,{\mathrm {e}}^{c\,x^2+b\,x+a}\,{\left (c\,x^2+b\,x+a\right )}^2+8\,{\mathrm {e}}^{c\,x^2+b\,x+a}\,{\left (c\,x^2+b\,x+a\right )}^3+8\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )\,{\left (-c\,x^2-b\,x-a\right )}^{7/2}}{15\,{\left (c\,x^2+b\,x+a\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(a + b*x + c*x^2)*(6*a + 6*b*x + 6*c*x^2) + 4*exp(a + b*x + c*x^2)*(a + b*x + c*x^2)^2 + 8*exp(a + b*x +
c*x^2)*(a + b*x + c*x^2)^3 + 8*pi^(1/2)*erfc((- a - b*x - c*x^2)^(1/2))*(- a - b*x - c*x^2)^(7/2))/(15*(a + b*
x + c*x^2)^(7/2))

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