3.59 \(\int (f x)^m \log (c (d+\frac{e}{x^2})^p) \, dx\)

Optimal. Leaf size=82 \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (m+1)}-\frac{2 e f p (f x)^{m-1} \, _2F_1\left (1,\frac{1-m}{2};\frac{3-m}{2};-\frac{e}{d x^2}\right )}{d \left (1-m^2\right )} \]

[Out]

(-2*e*f*p*(f*x)^(-1 + m)*Hypergeometric2F1[1, (1 - m)/2, (3 - m)/2, -(e/(d*x^2))])/(d*(1 - m^2)) + ((f*x)^(1 +
m)*Log[c*(d + e/x^2)^p])/(f*(1 + m))

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Rubi [A]  time = 0.0544225, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2455, 16, 339, 364} \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (m+1)}-\frac{2 e f p (f x)^{m-1} \, _2F_1\left (1,\frac{1-m}{2};\frac{3-m}{2};-\frac{e}{d x^2}\right )}{d \left (1-m^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e/x^2)^p],x]

[Out]

(-2*e*f*p*(f*x)^(-1 + m)*Hypergeometric2F1[1, (1 - m)/2, (3 - m)/2, -(e/(d*x^2))])/(d*(1 - m^2)) + ((f*x)^(1 +
m)*Log[c*(d + e/x^2)^p])/(f*(1 + m))

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (f x)^m \log \left (c \left (d+\frac{e}{x^2}\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (1+m)}+\frac{(2 e p) \int \frac{(f x)^{1+m}}{\left (d+\frac{e}{x^2}\right ) x^3} \, dx}{f (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (1+m)}+\frac{\left (2 e f^2 p\right ) \int \frac{(f x)^{-2+m}}{d+\frac{e}{x^2}} \, dx}{1+m}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (1+m)}-\frac{\left (2 e f p \left (\frac{1}{x}\right )^{-1+m} (f x)^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{x^{-m}}{d+e x^2} \, dx,x,\frac{1}{x}\right )}{1+m}\\ &=-\frac{2 e f p (f x)^{-1+m} \, _2F_1\left (1,\frac{1-m}{2};\frac{3-m}{2};-\frac{e}{d x^2}\right )}{d \left (1-m^2\right )}+\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )}{f (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0297495, size = 76, normalized size = 0.93 \[ \frac{(f x)^m \left (d (m-1) x^2 \log \left (c \left (d+\frac{e}{x^2}\right )^p\right )+2 e p \, _2F_1\left (1,\frac{1}{2}-\frac{m}{2};\frac{3}{2}-\frac{m}{2};-\frac{e}{d x^2}\right )\right )}{d (m-1) (m+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((f*x)^m*(2*e*p*Hypergeometric2F1[1, 1/2 - m/2, 3/2 - m/2, -(e/(d*x^2))] + d*(-1 + m)*x^2*Log[c*(d + e/x^2)^p]
))/(d*(-1 + m)*(1 + m)*x)

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Maple [F]  time = 3.868, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m}\ln \left ( c \left ( d+{\frac{e}{{x}^{2}}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*ln(c*(d+e/x^2)^p),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*(d+e/x^2)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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