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

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

[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

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Rubi [A]  time = 0.17, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2176, 2180, 2204} \[ -\frac {15 \sqrt {\pi } e^{5/2} F^{c \left (a-\frac {b d}{e}\right )} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{8 b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 e^2 \sqrt {d+e x} F^{c (a+b x)}}{4 b^3 c^3 \log ^3(F)}-\frac {5 e (d+e x)^{3/2} F^{c (a+b x)}}{2 b^2 c^2 \log ^2(F)}+\frac {(d+e x)^{5/2} F^{c (a+b x)}}{b c \log (F)} \]

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2180

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] &&  !$UseGamma === True

Rule 2204

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

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Mathematica [A]  time = 0.03, size = 72, normalized size = 0.42 \[ \frac {e^2 \sqrt {d+e x} F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {7}{2},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^3 c^3 \log ^3(F) \sqrt {-\frac {b c \log (F) (d+e x)}{e}}} \]

Antiderivative was successfully verified.

[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)])

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fricas [A]  time = 0.45, size = 167, normalized size = 0.97 \[ \frac {\frac {15 \, \sqrt {\pi } \sqrt {-\frac {b c \log \relax (F)}{e}} e^{3} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \relax (F)}{e}}\right )}{F^{\frac {b c d - a c e}{e}}} + 2 \, {\left (15 \, b c e^{2} \log \relax (F) + 4 \, {\left (b^{3} c^{3} e^{2} x^{2} + 2 \, b^{3} c^{3} d e x + b^{3} c^{3} d^{2}\right )} \log \relax (F)^{3} - 10 \, {\left (b^{2} c^{2} e^{2} x + b^{2} c^{2} d e\right )} \log \relax (F)^{2}\right )} \sqrt {e x + d} F^{b c x + a c}}{8 \, b^{4} c^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 1.30, size = 691, normalized size = 3.99 \[ -\frac {1}{8} \, {\left (\frac {8 \, \sqrt {\pi } d^{3} \operatorname {erf}\left (-\sqrt {-b c e \log \relax (F)} \sqrt {x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \relax (F) - a c e \log \relax (F)\right )} e^{\left (-1\right )} + 1\right )}}{\sqrt {-b c e \log \relax (F)}} - 12 \, d^{2} {\left (\frac {\sqrt {\pi } {\left (2 \, b c d \log \relax (F) + e\right )} \operatorname {erf}\left (-\sqrt {-b c e \log \relax (F)} \sqrt {x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \relax (F) - a c e \log \relax (F)\right )} e^{\left (-1\right )} + 1\right )}}{\sqrt {-b c e \log \relax (F)} b c \log \relax (F)} + \frac {2 \, \sqrt {x e + d} e^{\left ({\left ({\left (x e + d\right )} b c \log \relax (F) - b c d \log \relax (F) + a c e \log \relax (F)\right )} e^{\left (-1\right )} + 1\right )}}{b c \log \relax (F)}\right )} + 6 \, d {\left (\frac {\sqrt {\pi } {\left (4 \, b^{2} c^{2} d^{2} \log \relax (F)^{2} + 4 \, b c d e \log \relax (F) + 3 \, e^{2}\right )} \operatorname {erf}\left (-\sqrt {-b c e \log \relax (F)} \sqrt {x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \relax (F) - a c e \log \relax (F) + 2 \, e\right )} e^{\left (-1\right )} + 1\right )}}{\sqrt {-b c e \log \relax (F)} b^{2} c^{2} \log \relax (F)^{2}} - \frac {2 \, {\left (2 \, {\left (x e + d\right )}^{\frac {3}{2}} b c e \log \relax (F) - 4 \, \sqrt {x e + d} b c d e \log \relax (F) - 3 \, \sqrt {x e + d} e^{2}\right )} e^{\left ({\left ({\left (x e + d\right )} b c \log \relax (F) - b c d \log \relax (F) + a c e \log \relax (F) - 2 \, e\right )} e^{\left (-1\right )}\right )}}{b^{2} c^{2} \log \relax (F)^{2}}\right )} e^{2} - {\left (\frac {\sqrt {\pi } {\left (8 \, b^{3} c^{3} d^{3} \log \relax (F)^{3} + 12 \, b^{2} c^{2} d^{2} e \log \relax (F)^{2} + 18 \, b c d e^{2} \log \relax (F) + 15 \, e^{3}\right )} \operatorname {erf}\left (-\sqrt {-b c e \log \relax (F)} \sqrt {x e + d} e^{\left (-1\right )}\right ) e^{\left (-{\left (b c d \log \relax (F) - a c e \log \relax (F) + 3 \, e\right )} e^{\left (-1\right )} + 1\right )}}{\sqrt {-b c e \log \relax (F)} b^{3} c^{3} \log \relax (F)^{3}} + \frac {2 \, {\left (4 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} c^{2} e \log \relax (F)^{2} - 12 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} c^{2} d e \log \relax (F)^{2} + 12 \, \sqrt {x e + d} b^{2} c^{2} d^{2} e \log \relax (F)^{2} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} b c e^{2} \log \relax (F) + 18 \, \sqrt {x e + d} b c d e^{2} \log \relax (F) + 15 \, \sqrt {x e + d} e^{3}\right )} e^{\left ({\left ({\left (x e + d\right )} b c \log \relax (F) - b c d \log \relax (F) + a c e \log \relax (F) - 3 \, e\right )} e^{\left (-1\right )}\right )}}{b^{3} c^{3} \log \relax (F)^{3}}\right )} e^{3}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*sqrt(pi)*d^3*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*e^(-(b*c*d*log(F) - a*c*e*log(F))*e^(-1) +
1)/sqrt(-b*c*e*log(F)) - 12*d^2*(sqrt(pi)*(2*b*c*d*log(F) + e)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))
*e^(-(b*c*d*log(F) - a*c*e*log(F))*e^(-1) + 1)/(sqrt(-b*c*e*log(F))*b*c*log(F)) + 2*sqrt(x*e + d)*e^(((x*e + d
)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F))*e^(-1) + 1)/(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)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*e^(-(b*c*d*log(F) - a*c*e*log(F) +
2*e)*e^(-1) + 1)/(sqrt(-b*c*e*log(F))*b^2*c^2*log(F)^2) - 2*(2*(x*e + d)^(3/2)*b*c*e*log(F) - 4*sqrt(x*e + d)*
b*c*d*e*log(F) - 3*sqrt(x*e + d)*e^2)*e^(((x*e + d)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F) - 2*e)*e^(-1))/(b
^2*c^2*log(F)^2))*e^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)*erf(-sqrt(-b*c*e*log(F))*sqrt(x*e + d)*e^(-1))*e^(-(b*c*d*log(F) - a*c*e*log(F) + 3*e)*e^(-1) + 1)/(sq
rt(-b*c*e*log(F))*b^3*c^3*log(F)^3) + 2*(4*(x*e + d)^(5/2)*b^2*c^2*e*log(F)^2 - 12*(x*e + d)^(3/2)*b^2*c^2*d*e
*log(F)^2 + 12*sqrt(x*e + d)*b^2*c^2*d^2*e*log(F)^2 - 10*(x*e + d)^(3/2)*b*c*e^2*log(F) + 18*sqrt(x*e + d)*b*c
*d*e^2*log(F) + 15*sqrt(x*e + d)*e^3)*e^(((x*e + d)*b*c*log(F) - b*c*d*log(F) + a*c*e*log(F) - 3*e)*e^(-1))/(b
^3*c^3*log(F)^3))*e^3)*e^(-1)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{\frac {5}{2}} F^{\left (b x +a \right ) c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{\frac {5}{2}} F^{{\left (b x + a\right )} c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int F^{c\,\left (a+b\,x\right )}\,{\left (d+e\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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