3.50 \(\int (F^{c (a+b x)})^n (d+e x)^{4/3} \, dx\)

Optimal. Leaf size=98 \[ -\frac {e \sqrt [3]{d+e x} \left (F^{c (a+b x)}\right )^n F^{c n \left (a-\frac {b d}{e}\right )-c n (a+b x)} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n \log (F) (d+e x)}{e}}} \]

[Out]

-e*F^(c*(a-b*d/e)*n-c*n*(b*x+a))*(F^(c*(b*x+a)))^n*(e*x+d)^(1/3)*GAMMA(7/3,-b*c*n*(e*x+d)*ln(F)/e)/b^2/c^2/n^2
/ln(F)^2/(-b*c*n*(e*x+d)*ln(F)/e)^(1/3)

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Rubi [A]  time = 0.10, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2182, 2181} \[ -\frac {e \sqrt [3]{d+e x} \left (F^{c (a+b x)}\right )^n F^{c n \left (a-\frac {b d}{e}\right )-c n (a+b x)} \text {Gamma}\left (\frac {7}{3},-\frac {b c n \log (F) (d+e x)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n \log (F) (d+e x)}{e}}} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x]

[Out]

-((e*F^(c*(a - (b*d)/e)*n - c*n*(a + b*x))*(F^(c*(a + b*x)))^n*(d + e*x)^(1/3)*Gamma[7/3, -((b*c*n*(d + e*x)*L
og[F])/e)])/(b^2*c^2*n^2*Log[F]^2*(-((b*c*n*(d + e*x)*Log[F])/e))^(1/3)))

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2182

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*F^(g*(e +
f*x)))^n/F^(g*n*(e + f*x)), Int[(c + d*x)^m*F^(g*n*(e + f*x)), x], x] /; FreeQ[{F, b, c, d, e, f, g, m, n}, x]

Rubi steps

\begin {align*} \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx &=\left (F^{-c n (a+b x)} \left (F^{c (a+b x)}\right )^n\right ) \int F^{c n (a+b x)} (d+e x)^{4/3} \, dx\\ &=-\frac {e F^{c \left (a-\frac {b d}{e}\right ) n-c n (a+b x)} \left (F^{c (a+b x)}\right )^n \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n (d+e x) \log (F)}{e}}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 78, normalized size = 0.80 \[ -\frac {(d+e x)^{7/3} \left (F^{c (a+b x)}\right )^n F^{-\frac {b c n (d+e x)}{e}} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{e \left (-\frac {b c n \log (F) (d+e x)}{e}\right )^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x]

[Out]

-(((F^(c*(a + b*x)))^n*(d + e*x)^(7/3)*Gamma[7/3, -((b*c*n*(d + e*x)*Log[F])/e)])/(e*F^((b*c*n*(d + e*x))/e)*(
-((b*c*n*(d + e*x)*Log[F])/e))^(7/3)))

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fricas [A]  time = 0.49, size = 133, normalized size = 1.36 \[ \frac {\frac {4 \, \left (-\frac {b c n \log \relax (F)}{e}\right )^{\frac {2}{3}} e^{2} \Gamma \left (\frac {1}{3}, -\frac {{\left (b c e n x + b c d n\right )} \log \relax (F)}{e}\right )}{F^{\frac {{\left (b c d - a c e\right )} n}{e}}} - 3 \, {\left (4 \, b c e n \log \relax (F) - 3 \, {\left (b^{2} c^{2} e n^{2} x + b^{2} c^{2} d n^{2}\right )} \log \relax (F)^{2}\right )} {\left (e x + d\right )}^{\frac {1}{3}} F^{b c n x + a c n}}{9 \, b^{3} c^{3} n^{3} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^(c*(b*x+a)))^n*(e*x+d)^(4/3),x, algorithm="fricas")

[Out]

1/9*(4*(-b*c*n*log(F)/e)^(2/3)*e^2*gamma(1/3, -(b*c*e*n*x + b*c*d*n)*log(F)/e)/F^((b*c*d - a*c*e)*n/e) - 3*(4*
b*c*e*n*log(F) - 3*(b^2*c^2*e*n^2*x + b^2*c^2*d*n^2)*log(F)^2)*(e*x + d)^(1/3)*F^(b*c*n*x + a*c*n))/(b^3*c^3*n
^3*log(F)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{\frac {4}{3}} {\left (F^{{\left (b x + a\right )} c}\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^(c*(b*x+a)))^n*(e*x+d)^(4/3),x, algorithm="giac")

[Out]

integrate((e*x + d)^(4/3)*(F^((b*x + a)*c))^n, x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{\frac {4}{3}} \left (F^{\left (b x +a \right ) c}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x)

[Out]

int((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{\frac {4}{3}} {\left (F^{{\left (b x + a\right )} c}\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^(c*(b*x+a)))^n*(e*x+d)^(4/3),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(4/3)*(F^((b*x + a)*c))^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (F^{c\,\left (a+b\,x\right )}\right )}^n\,{\left (d+e\,x\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x)

[Out]

int((F^(c*(a + b*x)))^n*(d + e*x)^(4/3), 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)))**n*(e*x+d)**(4/3),x)

[Out]

Timed out

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