Integrand size = 19, antiderivative size = 165 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{15 e^2 (d+e x)^{3/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{15 e^3 \sqrt {d+e x}}+\frac {8 b^{5/2} c^{5/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right ) \log ^{\frac {5}{2}}(F)}{15 e^{7/2}} \] Output:
-2/5*F^(c*(b*x+a))/e/(e*x+d)^(5/2)-4/15*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d )^(3/2)-8/15*b^2*c^2*F^(c*(b*x+a))*ln(F)^2/e^3/(e*x+d)^(1/2)+8/15*b^(5/2)* c^(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))*ln(F)^(5/2)/e^(7/2)
Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.72 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=\frac {2 \left (-3 e^2 F^{c (a+b x)}-2 b c (d+e x) \log (F) \left (2 e F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{3/2}+F^{c (a+b x)} (e+2 b c (d+e x) \log (F))\right )\right )}{15 e^3 (d+e x)^{5/2}} \] Input:
Integrate[F^(c*(a + b*x))/(d + e*x)^(7/2),x]
Output:
(2*(-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]))))/(15*e^3*(d + e*x)^( 5/2))
Time = 0.72 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.
\(\displaystyle \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}}dx}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^{3/2}}dx}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 b c \log (F) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}}dx}{e}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {4 b c \log (F) \int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c (d+e x)}{e}}d\sqrt {d+e x}}{e^2}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2 b c \log (F) \left (\frac {2 b c \log (F) \left (\frac {2 \sqrt {\pi } \sqrt {b} \sqrt {c} \sqrt {\log (F)} F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 F^{c (a+b x)}}{e \sqrt {d+e x}}\right )}{3 e}-\frac {2 F^{c (a+b x)}}{3 e (d+e x)^{3/2}}\right )}{5 e}-\frac {2 F^{c (a+b x)}}{5 e (d+e x)^{5/2}}\) |
Input:
Int[F^(c*(a + b*x))/(d + e*x)^(7/2),x]
Output:
(-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)
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[$UseGamma]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : > Simp[2/d Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d *x]], x] /; FreeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
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]
\[\int \frac {F^{c \left (b x +a \right )}}{\left (e x +d \right )^{\frac {7}{2}}}d x\]
Input:
int(F^(c*(b*x+a))/(e*x+d)^(7/2),x)
Output:
int(F^(c*(b*x+a))/(e*x+d)^(7/2),x)
Time = 0.08 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.39 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {4 \, \sqrt {\pi } {\left (b^{2} c^{2} e^{3} x^{3} + 3 \, b^{2} c^{2} d e^{2} x^{2} + 3 \, b^{2} c^{2} d^{2} e x + b^{2} c^{2} d^{3}\right )} \sqrt {-\frac {b c \log \left (F\right )}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right ) \log \left (F\right )^{2}}{F^{\frac {b c d - a c e}{e}}} + {\left (4 \, {\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} \log \left (F\right )^{2} + 3 \, e^{2} + 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} \sqrt {e x + d} F^{b c x + a c}\right )}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \] Input:
integrate(F^((b*x+a)*c)/(e*x+d)^(7/2),x, algorithm="fricas")
Output:
-2/15*(4*sqrt(pi)*(b^2*c^2*e^3*x^3 + 3*b^2*c^2*d*e^2*x^2 + 3*b^2*c^2*d^2*e *x + b^2*c^2*d^3)*sqrt(-b*c*log(F)/e)*erf(sqrt(e*x + d)*sqrt(-b*c*log(F)/e ))*log(F)^2/F^((b*c*d - a*c*e)/e) + (4*(b^2*c^2*e^2*x^2 + 2*b^2*c^2*d*e*x + b^2*c^2*d^2)*log(F)^2 + 3*e^2 + 2*(b*c*e^2*x + b*c*d*e)*log(F))*sqrt(e*x + d)*F^(b*c*x + a*c))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)
\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate(F**((b*x+a)*c)/(e*x+d)**(7/2),x)
Output:
Integral(F**(c*(a + b*x))/(d + e*x)**(7/2), x)
\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(F^((b*x+a)*c)/(e*x+d)^(7/2),x, algorithm="maxima")
Output:
integrate(F^((b*x + a)*c)/(e*x + d)^(7/2), x)
\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(F^((b*x+a)*c)/(e*x+d)^(7/2),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)/(e*x + d)^(7/2), x)
Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{7/2}} \,d x \] Input:
int(F^(c*(a + b*x))/(d + e*x)^(7/2),x)
Output:
int(F^(c*(a + b*x))/(d + e*x)^(7/2), x)
\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx=f^{a c} \left (\int \frac {f^{b c x}}{\sqrt {e x +d}\, d^{3}+3 \sqrt {e x +d}\, d^{2} e x +3 \sqrt {e x +d}\, d \,e^{2} x^{2}+\sqrt {e x +d}\, e^{3} x^{3}}d x \right ) \] Input:
int(F^((b*x+a)*c)/(e*x+d)^(7/2),x)
Output:
f**(a*c)*int(f**(b*c*x)/(sqrt(d + e*x)*d**3 + 3*sqrt(d + e*x)*d**2*e*x + 3 *sqrt(d + e*x)*d*e**2*x**2 + sqrt(d + e*x)*e**3*x**3),x)