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

Optimal. Leaf size=81 \[ \frac{(f x)^{m+1} \log \left (c \left (d+e x^2\right )^p\right )}{f (m+1)}-\frac{2 e p (f x)^{m+3} \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )}{d f^3 (m+1) (m+3)} \]

[Out]

(-2*e*p*(f*x)^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)])/(d*f^3*(1 + m)*(3 + m)) + ((f*
x)^(1 + m)*Log[c*(d + e*x^2)^p])/(f*(1 + m))

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Rubi [A]  time = 0.0418747, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2455, 16, 364} \[ \frac{(f x)^{m+1} \log \left (c \left (d+e x^2\right )^p\right )}{f (m+1)}-\frac{2 e p (f x)^{m+3} \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )}{d f^3 (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e*x^2)^p],x]

[Out]

(-2*e*p*(f*x)^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)])/(d*f^3*(1 + m)*(3 + m)) + ((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 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+e x^2\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)}-\frac{(2 e p) \int \frac{x (f x)^{1+m}}{d+e x^2} \, dx}{f (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)}-\frac{(2 e p) \int \frac{(f x)^{2+m}}{d+e x^2} \, dx}{f^2 (1+m)}\\ &=-\frac{2 e p (f x)^{3+m} \, _2F_1\left (1,\frac{3+m}{2};\frac{5+m}{2};-\frac{e x^2}{d}\right )}{d f^3 (1+m) (3+m)}+\frac{(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0270598, size = 70, normalized size = 0.86 \[ \frac{x (f x)^m \left (d (m+3) \log \left (c \left (d+e x^2\right )^p\right )-2 e p x^2 \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{e x^2}{d}\right )\right )}{d (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(f*x)^m*(-2*e*p*x^2*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((e*x^2)/d)] + d*(3 + m)*Log[c*(d + e*x^2)^
p]))/(d*(1 + m)*(3 + m))

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Maple [F]  time = 1.053, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m}\ln \left ( c \left ( e{x}^{2}+d \right ) ^{p} \right ) \, 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),x)

[Out]

int((f*x)^m*ln(c*(e*x^2+d)^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*(e*x^2+d)^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 ({\left (e x^{2} + d\right )}^{p} c\right ), 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),x, algorithm="fricas")

[Out]

integral((f*x)^m*log((e*x^2 + d)^p*c), 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),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 ({\left (e x^{2} + d\right )}^{p} c\right )\,{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),x, algorithm="giac")

[Out]

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