\(\int F^{c (a+b x)} (d+e x)^{5/2} \, dx\) [40]

Optimal result

Integrand size = 19, antiderivative size = 173 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=-\frac {15 e^{5/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 e^2 F^{c (a+b x)} \sqrt {d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)} \]

[Out]

-5/2*e*F^(c*(b*x+a))*(e*x+d)^(3/2)/b^2/c^2/ln(F)^2+F^(c*(b*x+a))*(e*x+d)^(5/2)/b/c/ln(F)-15/8*e^(5/2)*F^(c*(a-
b*d/e))*erfi(b^(1/2)*c^(1/2)*(e*x+d)^(1/2)*ln(F)^(1/2)/e^(1/2))*Pi^(1/2)/b^(7/2)/c^(7/2)/ln(F)^(7/2)+15/4*e^2*
F^(c*(b*x+a))*(e*x+d)^(1/2)/b^3/c^3/ln(F)^3

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2207, 2211, 2235} \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=-\frac {15 \sqrt {\pi } e^{5/2} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 e^2 \sqrt {d+e x} F^{c (a+b x)}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e (d+e x)^{3/2} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)} \]

[In]

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

[Out]

(-15*e^(5/2)*F^(c*(a - (b*d)/e))*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(8*b^(7/
2)*c^(7/2)*Log[F]^(7/2)) + (15*e^2*F^(c*(a + b*x))*Sqrt[d + e*x])/(4*b^3*c^3*Log[F]^3) - (5*e*F^(c*(a + b*x))*
(d + e*x)^(3/2))/(2*b^2*c^2*Log[F]^2) + (F^(c*(a + b*x))*(d + e*x)^(5/2))/(b*c*Log[F])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac {(5 e) \int F^{c (a+b x)} (d+e x)^{3/2} \, dx}{2 b c \log (F)} \\ & = -\frac {5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}+\frac {\left (15 e^2\right ) \int F^{c (a+b x)} \sqrt {d+e x} \, dx}{4 b^2 c^2 \log ^2(F)} \\ & = \frac {15 e^2 F^{c (a+b x)} \sqrt {d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac {\left (15 e^3\right ) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}} \, dx}{8 b^3 c^3 \log ^3(F)} \\ & = \frac {15 e^2 F^{c (a+b x)} \sqrt {d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)}-\frac {\left (15 e^2\right ) \text {Subst}\left (\int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b^3 c^3 \log ^3(F)} \\ & = -\frac {15 e^{5/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 e^2 F^{c (a+b x)} \sqrt {d+e x}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e F^{c (a+b x)} (d+e x)^{3/2}}{2 b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^{5/2}}{b c \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.42 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=\frac {e^2 F^{c \left (a-\frac {b d}{e}\right )} \sqrt {d+e x} \Gamma \left (\frac {7}{2},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^3 c^3 \log ^3(F) \sqrt {-\frac {b c (d+e x) \log (F)}{e}}} \]

[In]

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

[Out]

(e^2*F^(c*(a - (b*d)/e))*Sqrt[d + e*x]*Gamma[7/2, -((b*c*(d + e*x)*Log[F])/e)])/(b^3*c^3*Log[F]^3*Sqrt[-((b*c*
(d + e*x)*Log[F])/e)])

Maple [F]

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

[In]

int(F^(c*(b*x+a))*(e*x+d)^(5/2),x)

[Out]

int(F^(c*(b*x+a))*(e*x+d)^(5/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {\pi } \sqrt {-\frac {b c \log \left (F\right )}{e}} e^{3} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right )}{F^{\frac {b c d - a c e}{e}}} + 2 \, {\left (15 \, b c e^{2} \log \left (F\right ) + 4 \, {\left (b^{3} c^{3} e^{2} x^{2} + 2 \, b^{3} c^{3} d e x + b^{3} c^{3} d^{2}\right )} \log \left (F\right )^{3} - 10 \, {\left (b^{2} c^{2} e^{2} x + b^{2} c^{2} d e\right )} \log \left (F\right )^{2}\right )} \sqrt {e x + d} F^{b c x + a c}}{8 \, b^{4} c^{4} \log \left (F\right )^{4}} \]

[In]

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

[Out]

1/8*(15*sqrt(pi)*sqrt(-b*c*log(F)/e)*e^3*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))/F^((b*c*d - a*c*e)/e) + 2*(15*
b*c*e^2*log(F) + 4*(b^3*c^3*e^2*x^2 + 2*b^3*c^3*d*e*x + b^3*c^3*d^2)*log(F)^3 - 10*(b^2*c^2*e^2*x + b^2*c^2*d*
e)*log(F)^2)*sqrt(e*x + d)*F^(b*c*x + a*c))/(b^4*c^4*log(F)^4)

Sympy [F]

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

[In]

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

[Out]

Integral(F**(c*(a + b*x))*(d + e*x)**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 643 vs. \(2 (141) = 282\).

Time = 0.43 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.72 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=-\frac {\frac {8 \, \sqrt {\pi } d^{3} e \operatorname {erf}\left (-\frac {\sqrt {-b c e \log \left (F\right )} \sqrt {e x + d}}{e}\right ) e^{\left (-\frac {b c d \log \left (F\right ) - a c e \log \left (F\right )}{e}\right )}}{\sqrt {-b c e \log \left (F\right )}} - 12 \, d^{2} {\left (\frac {\sqrt {\pi } {\left (2 \, b c d \log \left (F\right ) + e\right )} e \operatorname {erf}\left (-\frac {\sqrt {-b c e \log \left (F\right )} \sqrt {e x + d}}{e}\right ) e^{\left (-\frac {b c d \log \left (F\right ) - a c e \log \left (F\right )}{e}\right )}}{\sqrt {-b c e \log \left (F\right )} b c \log \left (F\right )} + \frac {2 \, \sqrt {e x + d} e e^{\left (\frac {{\left (e x + d\right )} b c \log \left (F\right ) - b c d \log \left (F\right ) + a c e \log \left (F\right )}{e}\right )}}{b c \log \left (F\right )}\right )} + 6 \, d {\left (\frac {\sqrt {\pi } {\left (4 \, b^{2} c^{2} d^{2} \log \left (F\right )^{2} + 4 \, b c d e \log \left (F\right ) + 3 \, e^{2}\right )} e \operatorname {erf}\left (-\frac {\sqrt {-b c e \log \left (F\right )} \sqrt {e x + d}}{e}\right ) e^{\left (-\frac {b c d \log \left (F\right ) - a c e \log \left (F\right )}{e}\right )}}{\sqrt {-b c e \log \left (F\right )} b^{2} c^{2} \log \left (F\right )^{2}} - \frac {2 \, {\left (2 \, {\left (e x + d\right )}^{\frac {3}{2}} b c e \log \left (F\right ) - 4 \, \sqrt {e x + d} b c d e \log \left (F\right ) - 3 \, \sqrt {e x + d} e^{2}\right )} e^{\left (\frac {{\left (e x + d\right )} b c \log \left (F\right ) - b c d \log \left (F\right ) + a c e \log \left (F\right )}{e}\right )}}{b^{2} c^{2} \log \left (F\right )^{2}}\right )} - \frac {\sqrt {\pi } {\left (8 \, b^{3} c^{3} d^{3} \log \left (F\right )^{3} + 12 \, b^{2} c^{2} d^{2} e \log \left (F\right )^{2} + 18 \, b c d e^{2} \log \left (F\right ) + 15 \, e^{3}\right )} e \operatorname {erf}\left (-\frac {\sqrt {-b c e \log \left (F\right )} \sqrt {e x + d}}{e}\right ) e^{\left (-\frac {b c d \log \left (F\right ) - a c e \log \left (F\right )}{e}\right )}}{\sqrt {-b c e \log \left (F\right )} b^{3} c^{3} \log \left (F\right )^{3}} - \frac {2 \, {\left (4 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} c^{2} e \log \left (F\right )^{2} - 12 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} c^{2} d e \log \left (F\right )^{2} + 12 \, \sqrt {e x + d} b^{2} c^{2} d^{2} e \log \left (F\right )^{2} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} b c e^{2} \log \left (F\right ) + 18 \, \sqrt {e x + d} b c d e^{2} \log \left (F\right ) + 15 \, \sqrt {e x + d} e^{3}\right )} e^{\left (\frac {{\left (e x + d\right )} b c \log \left (F\right ) - b c d \log \left (F\right ) + a c e \log \left (F\right )}{e}\right )}}{b^{3} c^{3} \log \left (F\right )^{3}}}{8 \, e} \]

[In]

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

[Out]

-1/8*(8*sqrt(pi)*d^3*e*erf(-sqrt(-b*c*e*log(F))*sqrt(e*x + d)/e)*e^(-(b*c*d*log(F) - a*c*e*log(F))/e)/sqrt(-b*
c*e*log(F)) - 12*d^2*(sqrt(pi)*(2*b*c*d*log(F) + e)*e*erf(-sqrt(-b*c*e*log(F))*sqrt(e*x + d)/e)*e^(-(b*c*d*log
(F) - a*c*e*log(F))/e)/(sqrt(-b*c*e*log(F))*b*c*log(F)) + 2*sqrt(e*x + d)*e*e^(((e*x + d)*b*c*log(F) - b*c*d*l
og(F) + a*c*e*log(F))/e)/(b*c*log(F))) + 6*d*(sqrt(pi)*(4*b^2*c^2*d^2*log(F)^2 + 4*b*c*d*e*log(F) + 3*e^2)*e*e
rf(-sqrt(-b*c*e*log(F))*sqrt(e*x + d)/e)*e^(-(b*c*d*log(F) - a*c*e*log(F))/e)/(sqrt(-b*c*e*log(F))*b^2*c^2*log
(F)^2) - 2*(2*(e*x + d)^(3/2)*b*c*e*log(F) - 4*sqrt(e*x + d)*b*c*d*e*log(F) - 3*sqrt(e*x + d)*e^2)*e^(((e*x +
d)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F))/e)/(b^2*c^2*log(F)^2)) - sqrt(pi)*(8*b^3*c^3*d^3*log(F)^3 + 12*b^
2*c^2*d^2*e*log(F)^2 + 18*b*c*d*e^2*log(F) + 15*e^3)*e*erf(-sqrt(-b*c*e*log(F))*sqrt(e*x + d)/e)*e^(-(b*c*d*lo
g(F) - a*c*e*log(F))/e)/(sqrt(-b*c*e*log(F))*b^3*c^3*log(F)^3) - 2*(4*(e*x + d)^(5/2)*b^2*c^2*e*log(F)^2 - 12*
(e*x + d)^(3/2)*b^2*c^2*d*e*log(F)^2 + 12*sqrt(e*x + d)*b^2*c^2*d^2*e*log(F)^2 - 10*(e*x + d)^(3/2)*b*c*e^2*lo
g(F) + 18*sqrt(e*x + d)*b*c*d*e^2*log(F) + 15*sqrt(e*x + d)*e^3)*e^(((e*x + d)*b*c*log(F) - b*c*d*log(F) + a*c
*e*log(F))/e)/(b^3*c^3*log(F)^3))/e

Mupad [F(-1)]

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

[In]

int(F^(c*(a + b*x))*(d + e*x)^(5/2),x)

[Out]

int(F^(c*(a + b*x))*(d + e*x)^(5/2), x)