Integrand size = 33, antiderivative size = 115 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}-\frac {4 e^{a+b x+c x^2}}{15 \left (a+b x+c x^2\right )^{3/2}}-\frac {8 e^{a+b x+c x^2}}{15 \sqrt {a+b x+c x^2}}+\frac {8}{15} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]
-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)
Time = 4.79 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.79 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {-2 e^{a+x (b+c x)} \left (3+2 (a+x (b+c x))+4 (a+x (b+c x))^2\right )+8 (-a-x (b+c x))^{5/2} \Gamma \left (\frac {1}{2},-a-x (b+c x)\right )}{15 (a+x (b+c x))^{5/2}} \]
(-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))
Time = 0.58 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7258, 2608, 2608, 2608, 2611, 2633}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(b+2 c x) e^{a+b x+c x^2}}{\left (a+b x+c x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 7258 |
\(\displaystyle \int \frac {e^{a+b x+c x^2}}{\left (a+b x+c x^2\right )^{7/2}}d\left (a+b x+c x^2\right )\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} \int \frac {e^{c x^2+b x+a}}{\left (c x^2+b x+a\right )^{5/2}}d\left (c x^2+b x+a\right )-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{3} \int \frac {e^{c x^2+b x+a}}{\left (c x^2+b x+a\right )^{3/2}}d\left (c x^2+b x+a\right )-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}\right )-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{3} \left (2 \int \frac {e^{c x^2+b x+a}}{\sqrt {c x^2+b x+a}}d\left (c x^2+b x+a\right )-\frac {2 e^{a+b x+c x^2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}\right )-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{3} \left (4 \int e^{c x^2+b x+a}d\sqrt {c x^2+b x+a}-\frac {2 e^{a+b x+c x^2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}\right )-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{3} \left (2 \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right )-\frac {2 e^{a+b x+c x^2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}\right )-\frac {2 e^{a+b x+c x^2}}{5 \left (a+b x+c x^2\right )^{5/2}}\) |
(-2*E^(a + b*x + c*x^2))/(5*(a + b*x + c*x^2)^(5/2)) + (2*((-2*E^(a + b*x + c*x^2))/(3*(a + b*x + c*x^2)^(3/2)) + (2*((-2*E^(a + b*x + c*x^2))/Sqrt[ a + b*x + c*x^2] + 2*Sqrt[Pi]*Erfi[Sqrt[a + b*x + c*x^2]]))/3))/5
3.7.34.3.1 Defintions of rubi rules used
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] - Simp[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] && In tegerQ[2*m] && !TrueQ[$UseGamma]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : > Simp[2/d Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d *x]], x] /; FreeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
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]
Int[(F_)^(v_)*(u_)*(w_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q Subst[Int[x^m*F^x, x], x, v], x] /; !FalseQ[q]] /; FreeQ[ {F, m}, x] && EqQ[w, v]
Time = 0.49 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{5 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{15}-\frac {8 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \sqrt {c \,x^{2}+b x +a}}\) | \(95\) |
default | \(-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{5 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {4 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {8 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{15}-\frac {8 \,{\mathrm e}^{c \,x^{2}+b x +a}}{15 \sqrt {c \,x^{2}+b x +a}}\) | \(95\) |
-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)
\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {7}{2}}} \,d x } \]
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)
Timed out. \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {7}{2}}} \,d x } \]
Time = 1.44 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {{\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (6\,c\,x^2+6\,b\,x+6\,a\right )+4\,{\mathrm {e}}^{c\,x^2+b\,x+a}\,{\left (c\,x^2+b\,x+a\right )}^2+8\,{\mathrm {e}}^{c\,x^2+b\,x+a}\,{\left (c\,x^2+b\,x+a\right )}^3+8\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )\,{\left (-c\,x^2-b\,x-a\right )}^{7/2}}{15\,{\left (c\,x^2+b\,x+a\right )}^{7/2}} \]