3.2.53 \(\int \frac {x (a+b \log (c x^n))}{(d+e x)^{3/2}} \, dx\) [153]

Optimal. Leaf size=94 \[ -\frac {4 b n \sqrt {d+e x}}{e^2}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{e^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2} \]

[Out]

8*b*n*arctanh((e*x+d)^(1/2)/d^(1/2))*d^(1/2)/e^2+2*d*(a+b*ln(c*x^n))/e^2/(e*x+d)^(1/2)-4*b*n*(e*x+d)^(1/2)/e^2
+2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^2

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Rubi [A]
time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {45, 2392, 12, 81, 65, 214} \begin {gather*} \frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}-\frac {4 b n \sqrt {d+e x}}{e^2}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*Log[c*x^n]))/(d + e*x)^(3/2),x]

[Out]

(-4*b*n*Sqrt[d + e*x])/e^2 + (8*b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/e^2 + (2*d*(a + b*Log[c*x^n]))/(e^
2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2392

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rubi steps

\begin {align*} \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx &=\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}-(b n) \int \frac {2 (2 d+e x)}{e^2 x \sqrt {d+e x}} \, dx\\ &=\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {(2 b n) \int \frac {2 d+e x}{x \sqrt {d+e x}} \, dx}{e^2}\\ &=-\frac {4 b n \sqrt {d+e x}}{e^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {(4 b d n) \int \frac {1}{x \sqrt {d+e x}} \, dx}{e^2}\\ &=-\frac {4 b n \sqrt {d+e x}}{e^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {(8 b d n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e^3}\\ &=-\frac {4 b n \sqrt {d+e x}}{e^2}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{e^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 83, normalized size = 0.88 \begin {gather*} \frac {2 \left (2 a d-2 b d n+a e x-2 b e n x+4 b \sqrt {d} n \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+b (2 d+e x) \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x)^(3/2),x]

[Out]

(2*(2*a*d - 2*b*d*n + a*e*x - 2*b*e*n*x + 4*b*Sqrt[d]*n*Sqrt[d + e*x]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + b*(2*d
+ e*x)*Log[c*x^n]))/(e^2*Sqrt[d + e*x])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e x +d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*ln(c*x^n))/(e*x+d)^(3/2),x)

[Out]

int(x*(a+b*ln(c*x^n))/(e*x+d)^(3/2),x)

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Maxima [A]
time = 0.51, size = 113, normalized size = 1.20 \begin {gather*} -4 \, {\left (\sqrt {d} e^{\left (-2\right )} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) + \sqrt {x e + d} e^{\left (-2\right )}\right )} b n + 2 \, {\left (\sqrt {x e + d} e^{\left (-2\right )} + \frac {d e^{\left (-2\right )}}{\sqrt {x e + d}}\right )} b \log \left (c x^{n}\right ) + 2 \, {\left (\sqrt {x e + d} e^{\left (-2\right )} + \frac {d e^{\left (-2\right )}}{\sqrt {x e + d}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

-4*(sqrt(d)*e^(-2)*log((sqrt(x*e + d) - sqrt(d))/(sqrt(x*e + d) + sqrt(d))) + sqrt(x*e + d)*e^(-2))*b*n + 2*(s
qrt(x*e + d)*e^(-2) + d*e^(-2)/sqrt(x*e + d))*b*log(c*x^n) + 2*(sqrt(x*e + d)*e^(-2) + d*e^(-2)/sqrt(x*e + d))
*a

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Fricas [A]
time = 0.39, size = 230, normalized size = 2.45 \begin {gather*} \left [\frac {2 \, {\left (2 \, {\left (b n x e + b d n\right )} \sqrt {d} \log \left (\frac {x e + 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (2 \, b d n + {\left (2 \, b n - a\right )} x e - 2 \, a d - {\left (b x e + 2 \, b d\right )} \log \left (c\right ) - {\left (b n x e + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{x e^{3} + d e^{2}}, -\frac {2 \, {\left (4 \, {\left (b n x e + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (2 \, b d n + {\left (2 \, b n - a\right )} x e - 2 \, a d - {\left (b x e + 2 \, b d\right )} \log \left (c\right ) - {\left (b n x e + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{x e^{3} + d e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

[2*(2*(b*n*x*e + b*d*n)*sqrt(d)*log((x*e + 2*sqrt(x*e + d)*sqrt(d) + 2*d)/x) - (2*b*d*n + (2*b*n - a)*x*e - 2*
a*d - (b*x*e + 2*b*d)*log(c) - (b*n*x*e + 2*b*d*n)*log(x))*sqrt(x*e + d))/(x*e^3 + d*e^2), -2*(4*(b*n*x*e + b*
d*n)*sqrt(-d)*arctan(sqrt(x*e + d)*sqrt(-d)/d) + (2*b*d*n + (2*b*n - a)*x*e - 2*a*d - (b*x*e + 2*b*d)*log(c) -
(b*n*x*e + 2*b*d*n)*log(x))*sqrt(x*e + d))/(x*e^3 + d*e^2)]

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Sympy [A]
time = 30.22, size = 153, normalized size = 1.63 \begin {gather*} \frac {\frac {2 a d}{\sqrt {d + e x}} + 2 a \sqrt {d + e x} - 2 b d \left (\frac {2 n \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} - \frac {\log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{\sqrt {d + e x}}\right ) + 2 b \left (\sqrt {d + e x} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )} - \frac {2 n \left (\frac {d e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + e \sqrt {d + e x}\right )}{e}\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*ln(c*x**n))/(e*x+d)**(3/2),x)

[Out]

(2*a*d/sqrt(d + e*x) + 2*a*sqrt(d + e*x) - 2*b*d*(2*n*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) - log(c*(-d/e + (d
+ e*x)/e)**n)/sqrt(d + e*x)) + 2*b*(sqrt(d + e*x)*log(c*(-d/e + (d + e*x)/e)**n) - 2*n*(d*e*atan(sqrt(d + e*x
)/sqrt(-d))/sqrt(-d) + e*sqrt(d + e*x))/e))/e**2

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Giac [A]
time = 2.31, size = 105, normalized size = 1.12 \begin {gather*} -\frac {8 \, b d n \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right ) e^{\left (-2\right )}}{\sqrt {-d}} + \frac {2 \, {\left ({\left (x e + d\right )} b n \log \left (x e\right ) + b d n \log \left (x e\right ) - 3 \, {\left (x e + d\right )} b n - b d n + {\left (x e + d\right )} b \log \left (c\right ) + b d \log \left (c\right ) + {\left (x e + d\right )} a + a d\right )} e^{\left (-2\right )}}{\sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

-8*b*d*n*arctan(sqrt(x*e + d)/sqrt(-d))*e^(-2)/sqrt(-d) + 2*((x*e + d)*b*n*log(x*e) + b*d*n*log(x*e) - 3*(x*e
+ d)*b*n - b*d*n + (x*e + d)*b*log(c) + b*d*log(c) + (x*e + d)*a + a*d)*e^(-2)/sqrt(x*e + d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*log(c*x^n)))/(d + e*x)^(3/2),x)

[Out]

int((x*(a + b*log(c*x^n)))/(d + e*x)^(3/2), x)

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