\(\int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx\) [75]

Optimal result

Integrand size = 19, antiderivative size = 130 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx=-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{3 e^2 \sqrt {d+e x}}+\frac {4 b^{3/2} c^{3/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right ) \log ^{\frac {3}{2}}(F)}{3 e^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.71 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (2 e F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{3/2}+F^{c (a+b x)} (e+2 b c (d+e x) \log (F))\right )}{3 e^2 (d+e x)^{3/2}} \] Input:

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

Output:

(-2*(2*e*F^(c*(a - (b*d)/e))*Gamma[1/2, -((b*c*(d + e*x)*Log[F])/e)]*(-((b 
*c*(d + e*x)*Log[F])/e))^(3/2) + F^(c*(a + b*x))*(e + 2*b*c*(d + e*x)*Log[ 
F])))/(3*e^2*(d + e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.

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

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{3/2}}dx}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}}dx}{e}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\)

\(\Big \downarrow \) 2611

\(\displaystyle \frac {2 b c \log (F) \left (\frac {4 b c \log (F) \int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c (d+e x)}{e}}d\sqrt {d+e x}}{e^2}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 \sqrt {\pi } \sqrt {b} \sqrt {c} \sqrt {\log (F)} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2608

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]
 

rule 2611

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]
 

rule 2633

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.08 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {2 \, \sqrt {\pi } {\left (b c e^{2} x^{2} + 2 \, b c d e x + b c d^{2}\right )} \sqrt {-\frac {b c \log \left (F\right )}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right ) \log \left (F\right )}{F^{\frac {b c d - a c e}{e}}} + \sqrt {e x + d} {\left (2 \, {\left (b c e x + b c d\right )} \log \left (F\right ) + e\right )} F^{b c x + a c}\right )}}{3 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} \] Input:

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

Output:

-2/3*(2*sqrt(pi)*(b*c*e^2*x^2 + 2*b*c*d*e*x + b*c*d^2)*sqrt(-b*c*log(F)/e) 
*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e))*log(F)/F^((b*c*d - a*c*e)/e) + sqr 
t(e*x + d)*(2*(b*c*e*x + b*c*d)*log(F) + e)*F^(b*c*x + a*c))/(e^4*x^2 + 2* 
d*e^3*x + d^2*e^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)/(sqrt(d + e*x)*d**2 + 2*sqrt(d + e*x)*d*e*x + sqrt 
(d + e*x)*e**2*x**2),x)