3.60 \(\int (f x)^m \log (c (d+\frac {e}{x^3})^p) \, dx\)

Optimal. Leaf size=85 \[ \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}-\frac {3 e f^2 p (f x)^{m-2} \, _2F_1\left (1,\frac {2-m}{3};\frac {5-m}{3};-\frac {e}{d x^3}\right )}{d \left (-m^2+m+2\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, 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} \[ \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}-\frac {3 e f^2 p (f x)^{m-2} \, _2F_1\left (1,\frac {2-m}{3};\frac {5-m}{3};-\frac {e}{d x^3}\right )}{d \left (-m^2+m+2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e/x^3)^p],x]

[Out]

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

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])

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+\frac {e}{x^3}\right )^p\right ) \, dx &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (1+m)}+\frac {(3 e p) \int \frac {(f x)^{1+m}}{\left (d+\frac {e}{x^3}\right ) x^4} \, dx}{f (1+m)}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (1+m)}+\frac {\left (3 e f^3 p\right ) \int \frac {(f x)^{-3+m}}{d+\frac {e}{x^3}} \, dx}{1+m}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (1+m)}-\frac {\left (3 e f^2 p \left (\frac {1}{x}\right )^{-2+m} (f x)^{-2+m}\right ) \operatorname {Subst}\left (\int \frac {x^{1-m}}{d+e x^3} \, dx,x,\frac {1}{x}\right )}{1+m}\\ &=-\frac {3 e f^2 p (f x)^{-2+m} \, _2F_1\left (1,\frac {2-m}{3};\frac {5-m}{3};-\frac {e}{d x^3}\right )}{d \left (2+m-m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 76, normalized size = 0.89 \[ \frac {(f x)^m \left (d (m-2) x^3 \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )+3 e p \, _2F_1\left (1,\frac {2}{3}-\frac {m}{3};\frac {5}{3}-\frac {m}{3};-\frac {e}{d x^3}\right )\right )}{d (m-2) (m+1) x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*Log[c*(d + e/x^3)^p],x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (f x\right )^{m} \log \left (c \left (\frac {d x^{3} + e}{x^{3}}\right )^{p}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="fricas")

[Out]

integral((f*x)^m*log(c*((d*x^3 + e)/x^3)^p), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{3}}\right )}^{p}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="giac")

[Out]

integrate((f*x)^m*log(c*(d + e/x^3)^p), x)

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maple [F]  time = 8.85, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{x^{3}}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*ln(c*(d+e/x^3)^p),x)

[Out]

int((f*x)^m*ln(c*(d+e/x^3)^p),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {f^{m} x x^{m} \log \left ({\left (d x^{3} + e\right )}^{p}\right ) - 3 \, f^{m} x x^{m} \log \left (x^{p}\right )}{m + 1} + \int \frac {{\left (d f^{m} {\left (m + 1\right )} x^{3} \log \relax (c) + e f^{m} {\left (m + 1\right )} \log \relax (c) + 3 \, e f^{m} p\right )} x^{m}}{d {\left (m + 1\right )} x^{3} + e {\left (m + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="maxima")

[Out]

(f^m*x*x^m*log((d*x^3 + e)^p) - 3*f^m*x*x^m*log(x^p))/(m + 1) + integrate((d*f^m*(m + 1)*x^3*log(c) + e*f^m*(m
+ 1)*log(c) + 3*e*f^m*p)*x^m/(d*(m + 1)*x^3 + e*(m + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (c\,{\left (d+\frac {e}{x^3}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e/x^3)^p)*(f*x)^m,x)

[Out]

int(log(c*(d + e/x^3)^p)*(f*x)^m, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*ln(c*(d+e/x**3)**p),x)

[Out]

Timed out

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