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

Optimal. Leaf size=75 \[ \frac {(f x)^{m+1} \log ^2\left (c \left (d+e x^2\right )^p\right )}{f (m+1)}-\frac {4 e p \text {Int}\left (\frac {(f x)^{m+2} \log \left (c \left (d+e x^2\right )^p\right )}{d+e x^2},x\right )}{f^2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.09, 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 = {} \[ \int (f x)^m \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx \]

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*}

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Mathematica [A]  time = 1.08, size = 466, normalized size = 6.21 \[ \frac {(f x)^m \left (\frac {4 d (m+1) p^2 \left (\frac {e x^2}{d+e x^2}\right )^{\frac {1}{2}-\frac {m}{2}} \left ((m-1) \log \left (d+e x^2\right ) \, _2F_1\left (\frac {1}{2}-\frac {m}{2},\frac {1}{2}-\frac {m}{2};\frac {3}{2}-\frac {m}{2};\frac {d}{e x^2+d}\right )-2 \, _3F_2\left (\frac {1}{2}-\frac {m}{2},\frac {1}{2}-\frac {m}{2},\frac {1}{2}-\frac {m}{2};\frac {3}{2}-\frac {m}{2},\frac {3}{2}-\frac {m}{2};\frac {d}{e x^2+d}\right )\right )}{e (m-1)^2 x}+\frac {2 p \left (p \log \left (d+e x^2\right )-\log \left (c \left (d+e x^2\right )^p\right )\right ) \left (2 e x^3 \, _2F_1\left (1,\frac {m+3}{2};\frac {m+5}{2};-\frac {e x^2}{d}\right )-d (m+3) x \log \left (d+e x^2\right )\right )}{d (m+3)}-\frac {2 m p \left (p \log \left (d+e x^2\right )-\log \left (c \left (d+e x^2\right )^p\right )\right ) \left (d (m+3) x \log \left (d+e x^2\right )-2 e x^3 \, _2F_1\left (1,\frac {m+3}{2};\frac {m+5}{2};-\frac {e x^2}{d}\right )\right )}{d (m+3)}+m x \left (\log \left (c \left (d+e x^2\right )^p\right )-p \log \left (d+e x^2\right )\right )^2+x \left (\log \left (c \left (d+e x^2\right )^p\right )-p \log \left (d+e x^2\right )\right )^2+4 p^2 x \left (\frac {2 e x^2 \, _2F_1\left (1,\frac {m+3}{2};\frac {m+5}{2};-\frac {e x^2}{d}\right )}{d (m+3)}-\log \left (d+e x^2\right )\right )+(m+1) p^2 x \log ^2\left (d+e x^2\right )\right )}{(m+1)^2} \]

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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fricas [A]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2}, x\right ) \]

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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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2}\,{d x} \]

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)

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maple [A]  time = 1.27, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )^{2}\, dx \]

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 [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {f^{m} p^{2} x x^{m} \log \left (e x^{2} + d\right )^{2}}{m + 1} + \int \frac {2 \, {\left ({\left (m p + p\right )} d f^{m} \log \relax (c) - {\left (2 \, e f^{m} p^{2} - {\left (m p + p\right )} e f^{m} \log \relax (c)\right )} x^{2}\right )} x^{m} \log \left (e x^{2} + d\right ) + {\left (e f^{m} {\left (m + 1\right )} x^{2} \log \relax (c)^{2} + d f^{m} {\left (m + 1\right )} \log \relax (c)^{2}\right )} x^{m}}{e {\left (m + 1\right )} x^{2} + d {\left (m + 1\right )}}\,{d x} \]

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]

f^m*p^2*x*x^m*log(e*x^2 + d)^2/(m + 1) + integrate((2*((m*p + p)*d*f^m*log(c) - (2*e*f^m*p^2 - (m*p + p)*e*f^m
*log(c))*x^2)*x^m*log(e*x^2 + d) + (e*f^m*(m + 1)*x^2*log(c)^2 + d*f^m*(m + 1)*log(c)^2)*x^m)/(e*(m + 1)*x^2 +
d*(m + 1)), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}^2\,{\left (f\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^2)^p)^2*(f*x)^m,x)

[Out]

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

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}^{2}\, dx \]

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]

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

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