\(\int e^{a+b x+c x^2} (b+2 c x) (a+b x+c x^2)^{5/2} \, dx\) [628]

Optimal result

Integrand size = 33, antiderivative size = 112 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {15}{4} e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {5}{2} e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}+e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {15}{8} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6839, 2207, 2211, 2235} \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {15}{8} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right )+e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{5/2}-\frac {5}{2} e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}+\frac {15}{4} e^{a+b x+c x^2} \sqrt {a+b x+c x^2} \]

[In]

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

[Out]

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

Rule 2207

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] - Dist[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]

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

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.41 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\sqrt {a+x (b+c x)} \Gamma \left (\frac {7}{2},-a-x (b+c x)\right )}{\sqrt {-a-x (b+c x)}} \]

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*Gamma[7/2, -a - x*(b + c*x)])/Sqrt[-a - x*(b + c*x)]

Maple [A] (verified)

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	\(-\frac {5 \,{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{2}+{\mathrm e}^{c \,x^{2}+b x +a} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}-\frac {15 \,\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{8}+\frac {15 \,{\mathrm e}^{c \,x^{2}+b x +a} \sqrt {c \,x^{2}+b x +a}}{4}\)	\(94\)

[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]

-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)

Fricas [F]

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

[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((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)

Sympy [F(-1)]

Timed out. \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {1}{4} \, {\left (4 \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} + 15 \, \sqrt {c x^{2} + b x + a}\right )} e^{\left (c x^{2} + b x + a\right )} - \frac {15}{8} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, \sqrt {c x^{2} + b x + a}\right ) \]

[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]

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))

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left ({\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (\frac {15\,\sqrt {-c\,x^2-b\,x-a}}{4}+\frac {5\,{\left (-c\,x^2-b\,x-a\right )}^{3/2}}{2}+{\left (-c\,x^2-b\,x-a\right )}^{5/2}\right )+\frac {15\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )}{8}\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (-c\,x^2-b\,x-a\right )}^{5/2}} \]

[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)*((15*(- a - b*x - c*x^2)^(1/2))/4 + (5*(- a - b*x - c*x^2)^(3/2))/2 + (- a - b*x - c*x^
2)^(5/2)) + (15*pi^(1/2)*erfc((- a - b*x - c*x^2)^(1/2)))/8)*(a + b*x + c*x^2)^(5/2))/(- a - b*x - c*x^2)^(5/2
)