\(\int (f x)^m \log (c (d+e x^3)^p) \, dx\) [55]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=-\frac {3 e p (f x)^{4+m} \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},-\frac {e x^3}{d}\right )}{d f^4 (1+m) (4+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^3\right )^p\right )}{f (1+m)} \]

[Out]

-3*e*p*(f*x)^(4+m)*hypergeom([1, 4/3+1/3*m],[7/3+1/3*m],-e*x^3/d)/d/f^4/(1+m)/(4+m)+(f*x)^(1+m)*ln(c*(e*x^3+d)
^p)/f/(1+m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2505, 16, 371} \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {(f x)^{m+1} \log \left (c \left (d+e x^3\right )^p\right )}{f (m+1)}-\frac {3 e p (f x)^{m+4} \operatorname {Hypergeometric2F1}\left (1,\frac {m+4}{3},\frac {m+7}{3},-\frac {e x^3}{d}\right )}{d f^4 (m+1) (m+4)} \]

[In]

Int[(f*x)^m*Log[c*(d + e*x^3)^p],x]

[Out]

(-3*e*p*(f*x)^(4 + m)*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, -((e*x^3)/d)])/(d*f^4*(1 + m)*(4 + m)) + ((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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2505

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*} \text {integral}& = \frac {(f x)^{1+m} \log \left (c \left (d+e x^3\right )^p\right )}{f (1+m)}-\frac {(3 e p) \int \frac {x^2 (f x)^{1+m}}{d+e x^3} \, dx}{f (1+m)} \\ & = \frac {(f x)^{1+m} \log \left (c \left (d+e x^3\right )^p\right )}{f (1+m)}-\frac {(3 e p) \int \frac {(f x)^{3+m}}{d+e x^3} \, dx}{f^3 (1+m)} \\ & = -\frac {3 e p (f x)^{4+m} \, _2F_1\left (1,\frac {4+m}{3};\frac {7+m}{3};-\frac {e x^3}{d}\right )}{d f^4 (1+m) (4+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^3\right )^p\right )}{f (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {x (f x)^m \left (-3 e p x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},-\frac {e x^3}{d}\right )+d (4+m) \log \left (c \left (d+e x^3\right )^p\right )\right )}{d (1+m) (4+m)} \]

[In]

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

[Out]

(x*(f*x)^m*(-3*e*p*x^3*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, -((e*x^3)/d)] + d*(4 + m)*Log[c*(d + e*x^3)^
p]))/(d*(1 + m)*(4 + m))

Maple [F]

\[\int \left (f x \right )^{m} \ln \left (c \left (e \,x^{3}+d \right )^{p}\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\int \ln \left (c\,{\left (e\,x^3+d\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \]

[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)