3.4.99 \(\int (f x)^q (a+b \log (c (d+e x^m)^n)) \, dx\) [399]

Optimal. Leaf size=92 \[ -\frac {b e m n x^{1+m} (f x)^q \, _2F_1\left (1,\frac {1+m+q}{m};\frac {1+2 m+q}{m};-\frac {e x^m}{d}\right )}{d (1+q) (1+m+q)}+\frac {(f x)^{1+q} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (1+q)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2505, 20, 371} \begin {gather*} \frac {(f x)^{q+1} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (q+1)}-\frac {b e m n x^{m+1} (f x)^q \, _2F_1\left (1,\frac {m+q+1}{m};\frac {2 m+q+1}{m};-\frac {e x^m}{d}\right )}{d (q+1) (m+q+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f*x)^q*(a + b*Log[c*(d + e*x^m)^n]),x]

[Out]

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

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

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Mathematica [A]
time = 0.04, size = 82, normalized size = 0.89 \begin {gather*} \frac {x (f x)^q \left (-b e m n x^m \, _2F_1\left (1,\frac {1+m+q}{m};\frac {1+2 m+q}{m};-\frac {e x^m}{d}\right )+d (1+m+q) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )\right )}{d (1+q) (1+m+q)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{q} \left (a +b \ln \left (c \left (d +e \,x^{m}\right )^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^q*(a+b*ln(c*(d+e*x^m)^n)),x)

[Out]

int((f*x)^q*(a+b*ln(c*(d+e*x^m)^n)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*f^q*m^2*n*integrate(x^q/((m*(q + 1) - q^2 - 2*q - 1)*d^2 + 2*(m*(q + 1) - q^2 - 2*q - 1)*d*e^(m*log(x) +
1) + (m*(q + 1) - q^2 - 2*q - 1)*e^(2*m*log(x) + 2)), x) - (((m*(q + 1) - q^2 - 2*q - 1)*d*f^q*x + (m*(q + 1)
- q^2 - 2*q - 1)*f^q*x*e^(m*log(x) + 1))*x^q*log((d + e^(m*log(x) + 1))^n) + (((m*(q + 1) - q^2 - 2*q - 1)*f^q
*log(c) - (m^2*n - m*n*(q + 1))*f^q)*x*e^(m*log(x) + 1) - (d*f^q*m^2*n - (m*(q + 1) - q^2 - 2*q - 1)*d*f^q*log
(c))*x)*x^q)/((q^3 - (q^2 + 2*q + 1)*m + 3*q^2 + 3*q + 1)*d + (q^3 - (q^2 + 2*q + 1)*m + 3*q^2 + 3*q + 1)*e^(m
*log(x) + 1)))*b + (f*x)^(q + 1)*a/(f*(q + 1))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x)^q*b*log((x^m*e + d)^n*c) + (f*x)^q*a, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (f x\right )^{q} \left (a + b \log {\left (c \left (d + e x^{m}\right )^{n} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**q*(a+b*ln(c*(d+e*x**m)**n)),x)

[Out]

Integral((f*x)**q*(a + b*log(c*(d + e*x**m)**n)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log((x^m*e + d)^n*c) + a)*(f*x)^q, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (f\,x\right )}^q\,\left (a+b\,\ln \left (c\,{\left (d+e\,x^m\right )}^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^q*(a + b*log(c*(d + e*x^m)^n)),x)

[Out]

int((f*x)^q*(a + b*log(c*(d + e*x^m)^n)), x)

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