3.4.90 \(\int \frac {F^{a+b (c+d x)^2}}{(e+f x)^3} \, dx\) [390]

Optimal. Leaf size=200 \[ -\frac {F^{a+b (c+d x)^2}}{2 f (e+f x)^2}+\frac {b d (d e-c f) F^{a+b (c+d x)^2} \log (F)}{f^3 (e+f x)}-\frac {b^{3/2} d^2 (d e-c f) F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {3}{2}}(F)}{f^4}+\frac {b d^2 \log (F) \text {Int}\left (\frac {F^{a+b (c+d x)^2}}{e+f x},x\right )}{f^2}+\frac {2 b^2 d^2 (d e-c f)^2 \log ^2(F) \text {Int}\left (\frac {F^{a+b (c+d x)^2}}{e+f x},x\right )}{f^4} \]

[Out]

-1/2*F^(a+b*(d*x+c)^2)/f/(f*x+e)^2+b*d*(-c*f+d*e)*F^(a+b*(d*x+c)^2)*ln(F)/f^3/(f*x+e)-b^(3/2)*d^2*(-c*f+d*e)*F
^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1/2))*ln(F)^(3/2)*Pi^(1/2)/f^4+b*d^2*ln(F)*Unintegrable(F^(a+b*(d*x+c)^2)/(f*x+
e),x)/f^2+2*b^2*d^2*(-c*f+d*e)^2*ln(F)^2*Unintegrable(F^(a+b*(d*x+c)^2)/(f*x+e),x)/f^4

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Rubi [A]
time = 0.21, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {F^{a+b (c+d x)^2}}{(e+f x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[F^(a + b*(c + d*x)^2)/(e + f*x)^3,x]

[Out]

-1/2*F^(a + b*(c + d*x)^2)/(f*(e + f*x)^2) + (b*d*(d*e - c*f)*F^(a + b*(c + d*x)^2)*Log[F])/(f^3*(e + f*x)) -
(b^(3/2)*d^2*(d*e - c*f)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(3/2))/f^4 + (b*d^2*Log[F]*D
efer[Int][F^(a + b*(c + d*x)^2)/(e + f*x), x])/f^2 + (2*b^2*d^2*(d*e - c*f)^2*Log[F]^2*Defer[Int][F^(a + b*(c
+ d*x)^2)/(e + f*x), x])/f^4

Rubi steps

\begin {align*} \int \frac {F^{a+b (c+d x)^2}}{(e+f x)^3} \, dx &=-\frac {F^{a+b (c+d x)^2}}{2 f (e+f x)^2}+\frac {\left (b d^2 \log (F)\right ) \int \frac {F^{a+b (c+d x)^2}}{e+f x} \, dx}{f^2}-\frac {(b d (d e-c f) \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(e+f x)^2} \, dx}{f^2}\\ &=-\frac {F^{a+b (c+d x)^2}}{2 f (e+f x)^2}+\frac {b d (d e-c f) F^{a+b (c+d x)^2} \log (F)}{f^3 (e+f x)}+\frac {\left (b d^2 \log (F)\right ) \int \frac {F^{a+b (c+d x)^2}}{e+f x} \, dx}{f^2}-\frac {\left (2 b^2 d^3 (d e-c f) \log ^2(F)\right ) \int F^{a+b (c+d x)^2} \, dx}{f^4}+\frac {\left (2 b^2 d^2 (d e-c f)^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{e+f x} \, dx}{f^4}\\ &=-\frac {F^{a+b (c+d x)^2}}{2 f (e+f x)^2}+\frac {b d (d e-c f) F^{a+b (c+d x)^2} \log (F)}{f^3 (e+f x)}-\frac {b^{3/2} d^2 (d e-c f) F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {3}{2}}(F)}{f^4}+\frac {\left (b d^2 \log (F)\right ) \int \frac {F^{a+b (c+d x)^2}}{e+f x} \, dx}{f^2}+\frac {\left (2 b^2 d^2 (d e-c f)^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{e+f x} \, dx}{f^4}\\ \end {align*}

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Mathematica [A]
time = 1.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{a+b (c+d x)^2}}{(e+f x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(e + f*x)^3,x]

[Out]

Integrate[F^(a + b*(c + d*x)^2)/(e + f*x)^3, x]

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Maple [A]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {F^{a +b \left (d x +c \right )^{2}}}{\left (f x +e \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)/(f*x+e)^3,x)

[Out]

int(F^(a+b*(d*x+c)^2)/(f*x+e)^3,x)

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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(f*x+e)^3,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(f*x + e)^3, x)

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Fricas [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(f*x+e)^3,x, algorithm="fricas")

[Out]

integral(F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)/(f^3*x^3 + 3*f^2*x^2*e + 3*f*x*e^2 + e^3), x)

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (e + f x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)/(f*x+e)**3,x)

[Out]

Integral(F**(a + b*(c + d*x)**2)/(e + f*x)**3, x)

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Giac [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(f*x+e)^3,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(f*x + e)^3, x)

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {F^{a+b\,{\left (c+d\,x\right )}^2}}{{\left (e+f\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^2)/(e + f*x)^3,x)

[Out]

int(F^(a + b*(c + d*x)^2)/(e + f*x)^3, x)

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