\(\int F^{c (a+b x)} (d+e x)^{5/2} \, dx\) [70]

Optimal result

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

-15/8*e^(5/2)*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))/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)^3-5/2*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)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.42 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=\frac {e^2 F^{c \left (a-\frac {b d}{e}\right )} \sqrt {d+e x} \Gamma \left (\frac {7}{2},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^3 c^3 \log ^3(F) \sqrt {-\frac {b c (d+e x) \log (F)}{e}}} \] Input:

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

Output:

(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)])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.

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

\(\Big \downarrow \) 2607

\(\displaystyle \frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)}-\frac {5 e \int F^{c (a+b x)} (d+e x)^{3/2}dx}{2 b c \log (F)}\)

\(\Big \downarrow \) 2607

\(\displaystyle \frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)}-\frac {5 e \left (\frac {(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)}-\frac {3 e \int F^{c (a+b x)} \sqrt {d+e x}dx}{2 b c \log (F)}\right )}{2 b c \log (F)}\)

\(\Big \downarrow \) 2607

\(\displaystyle \frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)}-\frac {5 e \left (\frac {(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)}-\frac {3 e \left (\frac {\sqrt {d+e x} F^{c (a+b x)}}{b c \log (F)}-\frac {e \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}}dx}{2 b c \log (F)}\right )}{2 b c \log (F)}\right )}{2 b c \log (F)}\)

\(\Big \downarrow \) 2611

\(\displaystyle \frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)}-\frac {5 e \left (\frac {(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)}-\frac {3 e \left (\frac {\sqrt {d+e x} F^{c (a+b x)}}{b c \log (F)}-\frac {\int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c (d+e x)}{e}}d\sqrt {d+e x}}{b c \log (F)}\right )}{2 b c \log (F)}\right )}{2 b c \log (F)}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)}-\frac {5 e \left (\frac {(d+e x)^{3/2} F^{c (a+b x)}}{b c \log (F)}-\frac {3 e \left (\frac {\sqrt {d+e x} F^{c (a+b x)}}{b c \log (F)}-\frac {\sqrt {\pi } \sqrt {e} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{2 b^{3/2} c^{3/2} \log ^{\frac {3}{2}}(F)}\right )}{2 b c \log (F)}\right )}{2 b c \log (F)}\)

Input:

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

Output:

(F^(c*(a + b*x))*(d + e*x)^(5/2))/(b*c*Log[F]) - (5*e*((F^(c*(a + b*x))*(d 
 + e*x)^(3/2))/(b*c*Log[F]) - (3*e*(-1/2*(Sqrt[e]*F^(c*(a - (b*d)/e))*Sqrt 
[Pi]*Erfi[(Sqrt[b]*Sqrt[c]*Sqrt[d + e*x]*Sqrt[Log[F]])/Sqrt[e]])/(b^(3/2)* 
c^(3/2)*Log[F]^(3/2)) + (F^(c*(a + b*x))*Sqrt[d + e*x])/(b*c*Log[F])))/(2* 
b*c*Log[F])))/(2*b*c*Log[F])
 

Defintions of rubi rules used

rule 2607

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]
 

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]

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

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 = 0.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {\pi } \sqrt {-\frac {b c \log \left (F\right )}{e}} e^{3} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right )}{F^{\frac {b c d - a c e}{e}}} + 2 \, {\left (15 \, b c e^{2} \log \left (F\right ) + 4 \, {\left (b^{3} c^{3} e^{2} x^{2} + 2 \, b^{3} c^{3} d e x + b^{3} c^{3} d^{2}\right )} \log \left (F\right )^{3} - 10 \, {\left (b^{2} c^{2} e^{2} x + b^{2} c^{2} d e\right )} \log \left (F\right )^{2}\right )} \sqrt {e x + d} F^{b c x + a c}}{8 \, b^{4} c^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

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]

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

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

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

Output:

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 = 0.18 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.72 \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-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*log(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*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^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*log(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*log(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*l 
og(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. \[ \int F^{c (a+b x)} (d+e x)^{5/2} \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d+e\,x\right )}^{5/2} \,d x \] 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]

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