3.238 \(\int \frac{h+i x}{(f+g x) (a+b \log (c (d+e x)^n))^2} \, dx\)

Optimal. Leaf size=142 \[ \frac{(g h-f i) \text{Unintegrable}\left (\frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2},x\right )}{g}+\frac{i e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e g n^2}-\frac{i (d+e x)}{b e g n \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

[Out]

(i*(d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b^2*e*E^(a/(b*n))*g*n^2*(c*(d + e*x)^n)^n^(-1))
 - (i*(d + e*x))/(b*e*g*n*(a + b*Log[c*(d + e*x)^n])) + ((g*h - f*i)*Unintegrable[1/((f + g*x)*(a + b*Log[c*(d
 + e*x)^n])^2), x])/g

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Rubi [A]  time = 0.196317, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{h+i x}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(h + i*x)/((f + g*x)*(a + b*Log[c*(d + e*x)^n])^2),x]

[Out]

(i*(d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b^2*e*E^(a/(b*n))*g*n^2*(c*(d + e*x)^n)^n^(-1))
 - (i*(d + e*x))/(b*e*g*n*(a + b*Log[c*(d + e*x)^n])) + ((g*h - f*i)*Defer[Int][1/((f + g*x)*(a + b*Log[c*(d +
 e*x)^n])^2), x])/g

Rubi steps

\begin{align*} \int \frac{h+238 x}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx &=\int \left (\frac{238}{g \left (a+b \log \left (c (d+e x)^n\right )\right )^2}+\frac{-238 f+g h}{g (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}\right ) \, dx\\ &=\frac{238 \int \frac{1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}+\frac{(-238 f+g h) \int \frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}\\ &=\frac{238 \operatorname{Subst}\left (\int \frac{1}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx,x,d+e x\right )}{e g}+\frac{(-238 f+g h) \int \frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}\\ &=-\frac{238 (d+e x)}{b e g n \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac{(-238 f+g h) \int \frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}+\frac{238 \operatorname{Subst}\left (\int \frac{1}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{b e g n}\\ &=-\frac{238 (d+e x)}{b e g n \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac{(-238 f+g h) \int \frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}+\frac{\left (238 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int \frac{e^{\frac{x}{n}}}{a+b x} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{b e g n^2}\\ &=\frac{238 e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e g n^2}-\frac{238 (d+e x)}{b e g n \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac{(-238 f+g h) \int \frac{1}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx}{g}\\ \end{align*}

Mathematica [A]  time = 1.19359, size = 0, normalized size = 0. \[ \int \frac{h+i x}{(f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(h + i*x)/((f + g*x)*(a + b*Log[c*(d + e*x)^n])^2),x]

[Out]

Integrate[(h + i*x)/((f + g*x)*(a + b*Log[c*(d + e*x)^n])^2), x]

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Maple [A]  time = 2.901, size = 0, normalized size = 0. \begin{align*} \int{\frac{ix+h}{ \left ( gx+f \right ) \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x+h)/(g*x+f)/(a+b*ln(c*(e*x+d)^n))^2,x)

[Out]

int((i*x+h)/(g*x+f)/(a+b*ln(c*(e*x+d)^n))^2,x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{e i x^{2} + d h +{\left (e h + d i\right )} x}{b^{2} e f n \log \left (c\right ) + a b e f n +{\left (b^{2} e g n \log \left (c\right ) + a b e g n\right )} x +{\left (b^{2} e g n x + b^{2} e f n\right )} \log \left ({\left (e x + d\right )}^{n}\right )} + \int \frac{e g i x^{2} + 2 \, e f i x + e f h -{\left (g h - f i\right )} d}{b^{2} e f^{2} n \log \left (c\right ) + a b e f^{2} n +{\left (b^{2} e g^{2} n \log \left (c\right ) + a b e g^{2} n\right )} x^{2} + 2 \,{\left (b^{2} e f g n \log \left (c\right ) + a b e f g n\right )} x +{\left (b^{2} e g^{2} n x^{2} + 2 \, b^{2} e f g n x + b^{2} e f^{2} n\right )} \log \left ({\left (e x + d\right )}^{n}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)/(g*x+f)/(a+b*log(c*(e*x+d)^n))^2,x, algorithm="maxima")

[Out]

-(e*i*x^2 + d*h + (e*h + d*i)*x)/(b^2*e*f*n*log(c) + a*b*e*f*n + (b^2*e*g*n*log(c) + a*b*e*g*n)*x + (b^2*e*g*n
*x + b^2*e*f*n)*log((e*x + d)^n)) + integrate((e*g*i*x^2 + 2*e*f*i*x + e*f*h - (g*h - f*i)*d)/(b^2*e*f^2*n*log
(c) + a*b*e*f^2*n + (b^2*e*g^2*n*log(c) + a*b*e*g^2*n)*x^2 + 2*(b^2*e*f*g*n*log(c) + a*b*e*f*g*n)*x + (b^2*e*g
^2*n*x^2 + 2*b^2*e*f*g*n*x + b^2*e*f^2*n)*log((e*x + d)^n)), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{i x + h}{a^{2} g x + a^{2} f +{\left (b^{2} g x + b^{2} f\right )} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \,{\left (a b g x + a b f\right )} \log \left ({\left (e x + d\right )}^{n} c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)/(g*x+f)/(a+b*log(c*(e*x+d)^n))^2,x, algorithm="fricas")

[Out]

integral((i*x + h)/(a^2*g*x + a^2*f + (b^2*g*x + b^2*f)*log((e*x + d)^n*c)^2 + 2*(a*b*g*x + a*b*f)*log((e*x +
d)^n*c)), x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{h + i x}{\left (a + b \log{\left (c \left (d + e x\right )^{n} \right )}\right )^{2} \left (f + g x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)/(g*x+f)/(a+b*ln(c*(e*x+d)**n))**2,x)

[Out]

Integral((h + i*x)/((a + b*log(c*(d + e*x)**n))**2*(f + g*x)), x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i x + h}{{\left (g x + f\right )}{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)/(g*x+f)/(a+b*log(c*(e*x+d)^n))^2,x, algorithm="giac")

[Out]

integrate((i*x + h)/((g*x + f)*(b*log((e*x + d)^n*c) + a)^2), x)