\(\int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx\) [77]

Optimal result

Integrand size = 19, antiderivative size = 200 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {16 b^{7/2} c^{7/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right ) \log ^{\frac {7}{2}}(F)}{105 e^{9/2}} \] Output:

-2/7*F^(c*(b*x+a))/e/(e*x+d)^(7/2)-4/35*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d 
)^(5/2)-8/105*b^2*c^2*F^(c*(b*x+a))*ln(F)^2/e^3/(e*x+d)^(3/2)-16/105*b^3*c 
^3*F^(c*(b*x+a))*ln(F)^3/e^4/(e*x+d)^(1/2)+16/105*b^(7/2)*c^(7/2)*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))*l 
n(F)^(7/2)/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.72 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\frac {2 \left (-15 e^3 F^{c (a+b x)}+2 b c (d+e x) \log (F) \left (-3 e^2 F^{c (a+b x)}-2 b c (d+e x) \log (F) \left (2 e F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{3/2}+F^{c (a+b x)} (e+2 b c (d+e x) \log (F))\right )\right )\right )}{105 e^4 (d+e x)^{7/2}} \] Input:

Integrate[F^(c*(a + b*x))/(d + e*x)^(9/2),x]
 

Output:

(2*(-15*e^3*F^(c*(a + b*x)) + 2*b*c*(d + e*x)*Log[F]*(-3*e^2*F^(c*(a + b*x 
)) - 2*b*c*(d + e*x)*Log[F]*(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])))))/(105*e^4*(d + e*x)^(7/2))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2608, 2608, 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.

\(\displaystyle \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}}dx}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}}dx}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\right )}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{3/2}}dx}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\right )}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}}dx}{e}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\right )}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

\(\Big \downarrow \) 2611

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {4 b c \log (F) \int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c (d+e x)}{e}}d\sqrt {d+e x}}{e^2}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\right )}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 \sqrt {\pi } \sqrt {b} \sqrt {c} \sqrt {\log (F)} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\right )}{7 e}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}\)

Input:

Int[F^(c*(a + b*x))/(d + e*x)^(9/2),x]
 

Output:

(-2*F^(c*(a + b*x)))/(7*e*(d + e*x)^(7/2)) + (2*b*c*Log[F]*((-2*F^(c*(a + 
b*x)))/(5*e*(d + e*x)^(5/2)) + (2*b*c*Log[F]*((-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)))/(5*e)))/(7 
*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]

\[\int \frac {F^{c \left (b x +a \right )}}{\left (e x +d \right )^{\frac {9}{2}}}d x\]

Input:

int(F^(c*(b*x+a))/(e*x+d)^(9/2),x)
 

Output:

int(F^(c*(b*x+a))/(e*x+d)^(9/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.60 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (\frac {8 \, \sqrt {\pi } {\left (b^{3} c^{3} e^{4} x^{4} + 4 \, b^{3} c^{3} d e^{3} x^{3} + 6 \, b^{3} c^{3} d^{2} e^{2} x^{2} + 4 \, b^{3} c^{3} d^{3} e x + b^{3} c^{3} d^{4}\right )} \sqrt {-\frac {b c \log \left (F\right )}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right ) \log \left (F\right )^{3}}{F^{\frac {b c d - a c e}{e}}} + {\left (8 \, {\left (b^{3} c^{3} e^{3} x^{3} + 3 \, b^{3} c^{3} d e^{2} x^{2} + 3 \, b^{3} c^{3} d^{2} e x + b^{3} c^{3} d^{3}\right )} \log \left (F\right )^{3} + 15 \, e^{3} + 4 \, {\left (b^{2} c^{2} e^{3} x^{2} + 2 \, b^{2} c^{2} d e^{2} x + b^{2} c^{2} d^{2} e\right )} \log \left (F\right )^{2} + 6 \, {\left (b c e^{3} x + b c d e^{2}\right )} \log \left (F\right )\right )} \sqrt {e x + d} F^{b c x + a c}\right )}}{105 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \] Input:

integrate(F^((b*x+a)*c)/(e*x+d)^(9/2),x, algorithm="fricas")
 

Output:

-2/105*(8*sqrt(pi)*(b^3*c^3*e^4*x^4 + 4*b^3*c^3*d*e^3*x^3 + 6*b^3*c^3*d^2* 
e^2*x^2 + 4*b^3*c^3*d^3*e*x + b^3*c^3*d^4)*sqrt(-b*c*log(F)/e)*erf(sqrt(e* 
x + d)*sqrt(-b*c*log(F)/e))*log(F)^3/F^((b*c*d - a*c*e)/e) + (8*(b^3*c^3*e 
^3*x^3 + 3*b^3*c^3*d*e^2*x^2 + 3*b^3*c^3*d^2*e*x + b^3*c^3*d^3)*log(F)^3 + 
 15*e^3 + 4*(b^2*c^2*e^3*x^2 + 2*b^2*c^2*d*e^2*x + b^2*c^2*d^2*e)*log(F)^2 
 + 6*(b*c*e^3*x + b*c*d*e^2)*log(F))*sqrt(e*x + d)*F^(b*c*x + a*c))/(e^8*x 
^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4*d^3*e^5*x + d^4*e^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\text {Timed out} \] Input:

integrate(F**((b*x+a)*c)/(e*x+d)**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate(F^((b*x+a)*c)/(e*x+d)^(9/2),x, algorithm="maxima")
 

Output:

integrate(F^((b*x + a)*c)/(e*x + d)^(9/2), x)
 

Giac [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate(F^((b*x+a)*c)/(e*x+d)^(9/2),x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)/(e*x + d)^(9/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{9/2}} \,d x \] Input:

int(F^(c*(a + b*x))/(d + e*x)^(9/2),x)
 

Output:

int(F^(c*(a + b*x))/(d + e*x)^(9/2), x)
 

Reduce [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=f^{a c} \left (\int \frac {f^{b c x}}{\sqrt {e x +d}\, d^{4}+4 \sqrt {e x +d}\, d^{3} e x +6 \sqrt {e x +d}\, d^{2} e^{2} x^{2}+4 \sqrt {e x +d}\, d \,e^{3} x^{3}+\sqrt {e x +d}\, e^{4} x^{4}}d x \right ) \] Input:

int(F^((b*x+a)*c)/(e*x+d)^(9/2),x)
 

Output:

f**(a*c)*int(f**(b*c*x)/(sqrt(d + e*x)*d**4 + 4*sqrt(d + e*x)*d**3*e*x + 6 
*sqrt(d + e*x)*d**2*e**2*x**2 + 4*sqrt(d + e*x)*d*e**3*x**3 + sqrt(d + e*x 
)*e**4*x**4),x)