Integrand size = 33, antiderivative size = 112 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {15}{4} e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {5}{2} e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}+e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {15}{8} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]
-5/2*exp(c*x^2+b*x+a)*(c*x^2+b*x+a)^(3/2)+exp(c*x^2+b*x+a)*(c*x^2+b*x+a)^( 5/2)-15/8*erfi((c*x^2+b*x+a)^(1/2))*Pi^(1/2)+15/4*exp(c*x^2+b*x+a)*(c*x^2+ b*x+a)^(1/2)
Time = 1.90 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.41 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\sqrt {a+x (b+c x)} \Gamma \left (\frac {7}{2},-a-x (b+c x)\right )}{\sqrt {-a-x (b+c x)}} \]
Time = 0.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7258, 2607, 2607, 2607, 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.
\(\displaystyle \int (b+2 c x) e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 7258 |
\(\displaystyle \int e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}d\left (a+b x+c x^2\right )\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} \int e^{c x^2+b x+a} \left (c x^2+b x+a\right )^{3/2}d\left (c x^2+b x+a\right )\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} \left (e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac {3}{2} \int e^{c x^2+b x+a} \sqrt {c x^2+b x+a}d\left (c x^2+b x+a\right )\right )\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} \left (e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac {3}{2} \left (e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \int \frac {e^{c x^2+b x+a}}{\sqrt {c x^2+b x+a}}d\left (c x^2+b x+a\right )\right )\right )\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} \left (e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac {3}{2} \left (e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\int e^{c x^2+b x+a}d\sqrt {c x^2+b x+a}\right )\right )\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} \left (e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac {3}{2} \left (e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right )\right )\right )\) |
E^(a + b*x + c*x^2)*(a + b*x + c*x^2)^(5/2) - (5*(E^(a + b*x + c*x^2)*(a + b*x + c*x^2)^(3/2) - (3*(E^(a + b*x + c*x^2)*Sqrt[a + b*x + c*x^2] - (Sqr t[Pi]*Erfi[Sqrt[a + b*x + c*x^2]])/2))/2))/2
3.7.28.3.1 Defintions of rubi rules used
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Simp[(c + d*x)^m*((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Simp[d*(m/(f*g*n*Log[F])) 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] && GtQ[m, 0] && IntegerQ[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 = 0.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(-\frac {5 \,{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{2}+{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}-\frac {15 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{8}+\frac {15 \,{\mathrm e}^{c \,x^{2}+b x +a} \sqrt {c \,x^{2}+b x +a}}{4}\) | \(94\) |
default | \(-\frac {5 \,{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{2}+{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}-\frac {15 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{8}+\frac {15 \,{\mathrm e}^{c \,x^{2}+b x +a} \sqrt {c \,x^{2}+b x +a}}{4}\) | \(94\) |
-5/2*exp(c*x^2+b*x+a)*(c*x^2+b*x+a)^(3/2)+exp(c*x^2+b*x+a)*(c*x^2+b*x+a)^( 5/2)-15/8*erfi((c*x^2+b*x+a)^(1/2))*Pi^(1/2)+15/4*exp(c*x^2+b*x+a)*(c*x^2+ b*x+a)^(1/2)
\[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )} \,d x } \]
integral((2*c^3*x^5 + 5*b*c^2*x^4 + 4*(b^2*c + a*c^2)*x^3 + a^2*b + (b^3 + 6*a*b*c)*x^2 + 2*(a*b^2 + a^2*c)*x)*sqrt(c*x^2 + b*x + a)*e^(c*x^2 + b*x + a), x)
Timed out. \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Timed out} \]
\[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )} \,d x } \]
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {1}{4} \, {\left (4 \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} + 15 \, \sqrt {c x^{2} + b x + a}\right )} e^{\left (c x^{2} + b x + a\right )} - \frac {15}{8} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, \sqrt {c x^{2} + b x + a}\right ) \]
1/4*(4*(c*x^2 + b*x + a)^(5/2) - 10*(c*x^2 + b*x + a)^(3/2) + 15*sqrt(c*x^ 2 + b*x + a))*e^(c*x^2 + b*x + a) - 15/8*I*sqrt(pi)*erf(-I*sqrt(c*x^2 + b* x + a))
Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left ({\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (\frac {15\,\sqrt {-c\,x^2-b\,x-a}}{4}+\frac {5\,{\left (-c\,x^2-b\,x-a\right )}^{3/2}}{2}+{\left (-c\,x^2-b\,x-a\right )}^{5/2}\right )+\frac {15\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )}{8}\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (-c\,x^2-b\,x-a\right )}^{5/2}} \]