3.159 \(\int (f x)^m \log ^2(c (d+e x^2)^p) \, dx\)

Optimal. Leaf size=74 \[ \frac{(f x)^{m+1} \log ^2\left (c \left (d+e x^2\right )^p\right )}{f (m+1)}-\frac{4 e p \text{Unintegrable}\left (\frac{(f x)^{m+2} \log \left (c \left (d+e x^2\right )^p\right )}{d+e x^2},x\right )}{f^2 (m+1)} \]

[Out]

((f*x)^(1 + m)*Log[c*(d + e*x^2)^p]^2)/(f*(1 + m)) - (4*e*p*Unintegrable[((f*x)^(2 + m)*Log[c*(d + e*x^2)^p])/
(d + e*x^2), x])/(f^2*(1 + m))

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Rubi [A]  time = 0.0889901, 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 = {} \[ \int (f x)^m \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(f*x)^m*Log[c*(d + e*x^2)^p]^2,x]

[Out]

((f*x)^(1 + m)*Log[c*(d + e*x^2)^p]^2)/(f*(1 + m)) - (4*e*p*Defer[Int][((f*x)^(2 + m)*Log[c*(d + e*x^2)^p])/(d
+ e*x^2), x])/(f^2*(1 + m))

Rubi steps

\begin{align*} \int (f x)^m \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log ^2\left (c \left (d+e x^2\right )^p\right )}{f (1+m)}-\frac{(4 e p) \int \frac{(f x)^{2+m} \log \left (c \left (d+e x^2\right )^p\right )}{d+e x^2} \, dx}{f^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 1.02958, size = 466, normalized size = 6.3 \[ \frac{(f x)^m \left (\frac{4 d (m+1) p^2 \left (\frac{e x^2}{d+e x^2}\right )^{\frac{1}{2}-\frac{m}{2}} \left ((m-1) \log \left (d+e x^2\right ) \, _2F_1\left (\frac{1}{2}-\frac{m}{2},\frac{1}{2}-\frac{m}{2};\frac{3}{2}-\frac{m}{2};\frac{d}{e x^2+d}\right )-2 \, _3F_2\left (\frac{1}{2}-\frac{m}{2},\frac{1}{2}-\frac{m}{2},\frac{1}{2}-\frac{m}{2};\frac{3}{2}-\frac{m}{2},\frac{3}{2}-\frac{m}{2};\frac{d}{e x^2+d}\right )\right )}{e (m-1)^2 x}+\frac{2 p \left (p \log \left (d+e x^2\right )-\log \left (c \left (d+e x^2\right )^p\right )\right ) \left (2 e x^3 \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )-d (m+3) x \log \left (d+e x^2\right )\right )}{d (m+3)}-\frac{2 m p \left (p \log \left (d+e x^2\right )-\log \left (c \left (d+e x^2\right )^p\right )\right ) \left (d (m+3) x \log \left (d+e x^2\right )-2 e x^3 \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )\right )}{d (m+3)}+m x \left (\log \left (c \left (d+e x^2\right )^p\right )-p \log \left (d+e x^2\right )\right )^2+x \left (\log \left (c \left (d+e x^2\right )^p\right )-p \log \left (d+e x^2\right )\right )^2+4 p^2 x \left (\frac{2 e x^2 \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )}{d (m+3)}-\log \left (d+e x^2\right )\right )+(m+1) p^2 x \log ^2\left (d+e x^2\right )\right )}{(m+1)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(f*x)^m*Log[c*(d + e*x^2)^p]^2,x]

[Out]

((f*x)^m*(4*p^2*x*((2*e*x^2*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)])/(d*(3 + m)) - Log[d + e*
x^2]) + (1 + m)*p^2*x*Log[d + e*x^2]^2 + (4*d*(1 + m)*p^2*((e*x^2)/(d + e*x^2))^(1/2 - m/2)*(-2*Hypergeometric
PFQ[{1/2 - m/2, 1/2 - m/2, 1/2 - m/2}, {3/2 - m/2, 3/2 - m/2}, d/(d + e*x^2)] + (-1 + m)*Hypergeometric2F1[1/2
- m/2, 1/2 - m/2, 3/2 - m/2, d/(d + e*x^2)]*Log[d + e*x^2]))/(e*(-1 + m)^2*x) + (2*p*(2*e*x^3*Hypergeometric2
F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)] - d*(3 + m)*x*Log[d + e*x^2])*(p*Log[d + e*x^2] - Log[c*(d + e*x^2)^
p]))/(d*(3 + m)) - (2*m*p*(-2*e*x^3*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)] + d*(3 + m)*x*Log
[d + e*x^2])*(p*Log[d + e*x^2] - Log[c*(d + e*x^2)^p]))/(d*(3 + m)) + x*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^
2)^p])^2 + m*x*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p])^2))/(1 + m)^2

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Maple [A]  time = 0.957, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( \ln \left ( c \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*ln(c*(e*x^2+d)^p)^2,x)

[Out]

int((f*x)^m*ln(c*(e*x^2+d)^p)^2,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(e*x^2+d)^p)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(e*x^2+d)^p)^2,x, algorithm="fricas")

[Out]

integral((f*x)^m*log((e*x^2 + d)^p*c)^2, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*ln(c*(e*x**2+d)**p)**2,x)

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(e*x^2+d)^p)^2,x, algorithm="giac")

[Out]

integrate((f*x)^m*log((e*x^2 + d)^p*c)^2, x)