3.146 \(\int \frac {x (a+b \log (c x^n))}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {8 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^2}+\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2} \]

[Out]

-4/9*b*n*(e*x+d)^(3/2)/e^2-8/3*b*d^(3/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^2+2/3*(e*x+d)^(3/2)*(a+b*ln(c*x^n)
)/e^2+8/3*b*d*n*(e*x+d)^(1/2)/e^2-2*d*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^2

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Rubi [A]  time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {43, 2350, 12, 80, 50, 63, 208} \[ -\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {8 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^2}+\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d*n*Sqrt[d + e*x])/(3*e^2) - (4*b*n*(d + e*x)^(3/2))/(9*e^2) - (8*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[
d]])/(3*e^2) - (2*d*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^2 + (2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*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 43

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 80

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 208

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

Rule 2350

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 )}{\sqrt {d+e x}} \, dx &=-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-(b n) \int \frac {2 (-2 d+e x) \sqrt {d+e x}}{3 e^2 x} \, dx\\ &=-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {(2 b n) \int \frac {(-2 d+e x) \sqrt {d+e x}}{x} \, dx}{3 e^2}\\ &=-\frac {4 b n (d+e x)^{3/2}}{9 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {(4 b d n) \int \frac {\sqrt {d+e x}}{x} \, dx}{3 e^2}\\ &=\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (4 b d^2 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{3 e^2}\\ &=\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (8 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 e^3}\\ &=\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2}-\frac {8 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 80, normalized size = 0.67 \[ -\frac {2 \left (\sqrt {d+e x} \left (6 a d-3 a e x+b (6 d-3 e x) \log \left (c x^n\right )-10 b d n+2 b e n x\right )+12 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(12*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(6*a*d - 10*b*d*n - 3*a*e*x + 2*b*e*n*x + b
*(6*d - 3*e*x)*Log[c*x^n])))/(9*e^2)

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fricas [A]  time = 0.46, size = 189, normalized size = 1.59 \[ \left [\frac {2 \, {\left (6 \, b d^{\frac {3}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (10 \, b d n - 6 \, a d - {\left (2 \, b e n - 3 \, a e\right )} x + 3 \, {\left (b e x - 2 \, b d\right )} \log \relax (c) + 3 \, {\left (b e n x - 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, e^{2}}, \frac {2 \, {\left (12 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (10 \, b d n - 6 \, a d - {\left (2 \, b e n - 3 \, a e\right )} x + 3 \, {\left (b e x - 2 \, b d\right )} \log \relax (c) + 3 \, {\left (b e n x - 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[2/9*(6*b*d^(3/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (10*b*d*n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3
*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*d*n)*log(x))*sqrt(e*x + d))/e^2, 2/9*(12*b*sqrt(-d)*d*n*arctan(sqrt
(e*x + d)*sqrt(-d)/d) + (10*b*d*n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*
d*n)*log(x))*sqrt(e*x + d))/e^2]

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giac [A]  time = 0.76, size = 145, normalized size = 1.22 \[ \frac {8 \, b d^{2} n \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right ) e^{\left (-2\right )}}{3 \, \sqrt {-d}} + \frac {2}{9} \, {\left (3 \, {\left (x e + d\right )}^{\frac {3}{2}} b n \log \left (x e\right ) - 9 \, \sqrt {x e + d} b d n \log \left (x e\right ) - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b n + 21 \, \sqrt {x e + d} b d n + 3 \, {\left (x e + d\right )}^{\frac {3}{2}} b \log \relax (c) - 9 \, \sqrt {x e + d} b d \log \relax (c) + 3 \, {\left (x e + d\right )}^{\frac {3}{2}} a - 9 \, \sqrt {x e + d} a d\right )} e^{\left (-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

8/3*b*d^2*n*arctan(sqrt(x*e + d)/sqrt(-d))*e^(-2)/sqrt(-d) + 2/9*(3*(x*e + d)^(3/2)*b*n*log(x*e) - 9*sqrt(x*e
+ d)*b*d*n*log(x*e) - 5*(x*e + d)^(3/2)*b*n + 21*sqrt(x*e + d)*b*d*n + 3*(x*e + d)^(3/2)*b*log(c) - 9*sqrt(x*e
+ d)*b*d*log(c) + 3*(x*e + d)^(3/2)*a - 9*sqrt(x*e + d)*a*d)*e^(-2)

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x}{\sqrt {e x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*ln(c*x^n)+a)/(e*x+d)^(1/2),x)

[Out]

int(x*(b*ln(c*x^n)+a)/(e*x+d)^(1/2),x)

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maxima [A]  time = 1.23, size = 127, normalized size = 1.07 \[ \frac {4}{9} \, b n {\left (\frac {3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{2}} - \frac {{\left (e x + d\right )}^{\frac {3}{2}} - 6 \, \sqrt {e x + d} d}{e^{2}}\right )} + \frac {2}{3} \, b {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{2}} - \frac {3 \, \sqrt {e x + d} d}{e^{2}}\right )} \log \left (c x^{n}\right ) + \frac {2}{3} \, a {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{2}} - \frac {3 \, \sqrt {e x + d} d}{e^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

4/9*b*n*(3*d^(3/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^2 - ((e*x + d)^(3/2) - 6*sqrt(e*
x + d)*d)/e^2) + 2/3*b*((e*x + d)^(3/2)/e^2 - 3*sqrt(e*x + d)*d/e^2)*log(c*x^n) + 2/3*a*((e*x + d)^(3/2)/e^2 -
3*sqrt(e*x + d)*d/e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*log(c*x^n)))/(d + e*x)^(1/2),x)

[Out]

int((x*(a + b*log(c*x^n)))/(d + e*x)^(1/2), 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(x*(a+b*ln(c*x**n))/(e*x+d)**(1/2),x)

[Out]

Timed out

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