3.71 \(\int (a+b (F^{g (e+f x)})^n)^2 (c+d x)^m \, dx\)

Optimal. Leaf size=228 \[ \frac {a^2 (c+d x)^{m+1}}{d (m+1)}+\frac {2 a b (c+d x)^m \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \left (-\frac {f g n \log (F) (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {f g n (c+d x) \log (F)}{d}\right )}{f g n \log (F)}+\frac {b^2 2^{-m-1} (c+d x)^m \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac {c f}{d}\right )-2 g n (e+f x)} \left (-\frac {f g n \log (F) (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 f g n (c+d x) \log (F)}{d}\right )}{f g n \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2183, 2182, 2181} \[ \frac {2 a b (c+d x)^m \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \left (-\frac {f g n \log (F) (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {f g n \log (F) (c+d x)}{d}\right )}{f g n \log (F)}+\frac {b^2 2^{-m-1} (c+d x)^m \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac {c f}{d}\right )-2 g n (e+f x)} \left (-\frac {f g n \log (F) (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 f g n \log (F) (c+d x)}{d}\right )}{f g n \log (F)}+\frac {a^2 (c+d x)^{m+1}}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x)^m,x]

[Out]

(a^2*(c + d*x)^(1 + m))/(d*(1 + m)) + (2^(-1 - m)*b^2*F^(2*(e - (c*f)/d)*g*n - 2*g*n*(e + f*x))*(F^(e*g + f*g*
x))^(2*n)*(c + d*x)^m*Gamma[1 + m, (-2*f*g*n*(c + d*x)*Log[F])/d])/(f*g*n*Log[F]*(-((f*g*n*(c + d*x)*Log[F])/d
))^m) + (2*a*b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n*(c + d*x)^m*Gamma[1 + m, -((f*g*n*(c
+ d*x)*Log[F])/d)])/(f*g*n*Log[F]*(-((f*g*n*(c + d*x)*Log[F])/d))^m)

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]

Rule 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x)^m \, dx &=\int \left (a^2 (c+d x)^m+2 a b \left (F^{e g+f g x}\right )^n (c+d x)^m+b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)^m\right ) \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+(2 a b) \int \left (F^{e g+f g x}\right )^n (c+d x)^m \, dx+b^2 \int \left (F^{e g+f g x}\right )^{2 n} (c+d x)^m \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\left (2 a b F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n\right ) \int F^{n (e g+f g x)} (c+d x)^m \, dx+\left (b^2 F^{-2 n (e g+f g x)} \left (F^{e g+f g x}\right )^{2 n}\right ) \int F^{2 n (e g+f g x)} (c+d x)^m \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {2^{-1-m} b^2 F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} (c+d x)^m \Gamma \left (1+m,-\frac {2 f g n (c+d x) \log (F)}{d}\right ) \left (-\frac {f g n (c+d x) \log (F)}{d}\right )^{-m}}{f g n \log (F)}+\frac {2 a b F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n (c+d x)^m \Gamma \left (1+m,-\frac {f g n (c+d x) \log (F)}{d}\right ) \left (-\frac {f g n (c+d x) \log (F)}{d}\right )^{-m}}{f g n \log (F)}\\ \end {align*}

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Mathematica [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x)^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x)^m,x]

[Out]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x)^m, x]

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fricas [A]  time = 0.45, size = 191, normalized size = 0.84 \[ \frac {4 \, {\left (a b d m + a b d\right )} e^{\left (\frac {{\left (d e - c f\right )} g n \log \relax (F) - d m \log \left (-\frac {f g n \log \relax (F)}{d}\right )}{d}\right )} \Gamma \left (m + 1, -\frac {{\left (d f g n x + c f g n\right )} \log \relax (F)}{d}\right ) + {\left (b^{2} d m + b^{2} d\right )} e^{\left (\frac {2 \, {\left (d e - c f\right )} g n \log \relax (F) - d m \log \left (-\frac {2 \, f g n \log \relax (F)}{d}\right )}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (d f g n x + c f g n\right )} \log \relax (F)}{d}\right ) + 2 \, {\left (a^{2} d f g n x + a^{2} c f g n\right )} {\left (d x + c\right )}^{m} \log \relax (F)}{2 \, {\left (d f g m + d f g\right )} n \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c)^m,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{2} {\left (d x + c\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c)^m,x, algorithm="giac")

[Out]

integrate(((F^((f*x + e)*g))^n*b + a)^2*(d*x + c)^m, x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (b \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right )^{2} \left (d x +c \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*(F^((f*x+e)*g))^n+a)^2*(d*x+c)^m,x)

[Out]

int((b*(F^((f*x+e)*g))^n+a)^2*(d*x+c)^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (F^{e g}\right )}^{2 \, n} b^{2} \int e^{\left (m \log \left (d x + c\right ) + 2 \, n \log \left (F^{f g x}\right )\right )}\,{d x} + 2 \, {\left (F^{e g}\right )}^{n} a b \int e^{\left (m \log \left (d x + c\right ) + n \log \left (F^{f g x}\right )\right )}\,{d x} + \frac {{\left (d x + c\right )}^{m + 1} a^{2}}{d {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c)^m,x, algorithm="maxima")

[Out]

(F^(e*g))^(2*n)*b^2*integrate(e^(m*log(d*x + c) + 2*n*log(F^(f*g*x))), x) + 2*(F^(e*g))^n*a*b*integrate(e^(m*l
og(d*x + c) + n*log(F^(f*g*x))), x) + (d*x + c)^(m + 1)*a^2/(d*(m + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x)^m,x)

[Out]

int((a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x)^m, 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((a+b*(F**(g*(f*x+e)))**n)**2*(d*x+c)**m,x)

[Out]

Timed out

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