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

Optimal. Leaf size=87 \[ \frac {(f x)^{m+1} \log \left (c \left (d+e x^n\right )^p\right )}{f (m+1)}-\frac {e n p x^{n+1} (f x)^m \, _2F_1\left (1,\frac {m+n+1}{n};\frac {m+2 n+1}{n};-\frac {e x^n}{d}\right )}{d (m+1) (m+n+1)} \]

[Out]

-e*n*p*x^(1+n)*(f*x)^m*hypergeom([1, (1+m+n)/n],[(1+m+2*n)/n],-e*x^n/d)/d/(1+m)/(1+m+n)+(f*x)^(1+m)*ln(c*(d+e*
x^n)^p)/f/(1+m)

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Rubi [A]  time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, 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, 20, 364} \[ \frac {(f x)^{m+1} \log \left (c \left (d+e x^n\right )^p\right )}{f (m+1)}-\frac {e n p x^{n+1} (f x)^m \, _2F_1\left (1,\frac {m+n+1}{n};\frac {m+2 n+1}{n};-\frac {e x^n}{d}\right )}{d (m+1) (m+n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e*x^n)^p],x]

[Out]

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

Rule 20

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.89 \[ \frac {x (f x)^m \left (d (m+n+1) \log \left (c \left (d+e x^n\right )^p\right )-e n p x^n \, _2F_1\left (1,\frac {m+n+1}{n};\frac {m+2 n+1}{n};-\frac {e x^n}{d}\right )\right )}{d (m+1) (m+n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (f x\right )^{m} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \ln \left (c \left (e \,x^{n}+d \right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^n)^p)*(f*x)^m,x)

[Out]

int(log(c*(d + e*x^n)^p)*(f*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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