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

Optimal. Leaf size=142 \[ -\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^2}+\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2} \]

[Out]

8/45*b*d*n*(e*x+d)^(3/2)/e^2-4/25*b*n*(e*x+d)^(5/2)/e^2-8/15*b*d^(5/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^2-2/
3*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^2+2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^2+8/15*b*d^2*n*(e*x+d)^(1/2)/e^2

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Rubi [A]  time = 0.10, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}-\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*n*Sqrt[d + e*x])/(15*e^2) + (8*b*d*n*(d + e*x)^(3/2))/(45*e^2) - (4*b*n*(d + e*x)^(5/2))/(25*e^2) - (
8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^2) - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2) + (2
*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*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 x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-(b n) \int \frac {2 (d+e x)^{3/2} (-2 d+3 e x)}{15 e^2 x} \, dx\\ &=-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {(2 b n) \int \frac {(d+e x)^{3/2} (-2 d+3 e x)}{x} \, dx}{15 e^2}\\ &=-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {(4 b d n) \int \frac {(d+e x)^{3/2}}{x} \, dx}{15 e^2}\\ &=\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (4 b d^2 n\right ) \int \frac {\sqrt {d+e x}}{x} \, dx}{15 e^2}\\ &=\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (4 b d^3 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{15 e^2}\\ &=\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (8 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{15 e^3}\\ &=\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 116, normalized size = 0.82 \[ \frac {2 \sqrt {d+e x} \left (15 a \left (-2 d^2+d e x+3 e^2 x^2\right )+15 b \left (-2 d^2+d e x+3 e^2 x^2\right ) \log \left (c x^n\right )+2 b n \left (31 d^2-8 d e x-9 e^2 x^2\right )\right )-120 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{225 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(2*b*n*(31*d^2 - 8*d*e*x - 9*e^2*x^2) + 15*
a*(-2*d^2 + d*e*x + 3*e^2*x^2) + 15*b*(-2*d^2 + d*e*x + 3*e^2*x^2)*Log[c*x^n]))/(225*e^2)

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fricas [A]  time = 0.55, size = 291, normalized size = 2.05 \[ \left [\frac {2 \, {\left (30 \, b d^{\frac {5}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - {\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \, {\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \relax (c) + 15 \, {\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{225 \, e^{2}}, \frac {2 \, {\left (60 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - {\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \, {\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \relax (c) + 15 \, {\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{225 \, e^{2}}\right ] \]

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]

[2/225*(30*b*d^(5/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (62*b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n -
5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e)*x + 15*(3*b*e^2*x^2 + b*d*e*x - 2*b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b
*d*e*n*x - 2*b*d^2*n)*log(x))*sqrt(e*x + d))/e^2, 2/225*(60*b*sqrt(-d)*d^2*n*arctan(sqrt(e*x + d)*sqrt(-d)/d)
+ (62*b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n - 5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e)*x + 15*(3*b*e^2*x^2 + b*d*e*x
- 2*b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b*d*e*n*x - 2*b*d^2*n)*log(x))*sqrt(e*x + d))/e^2]

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

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]

integrate(sqrt(e*x + d)*(b*log(c*x^n) + a)*x, x)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) \sqrt {e x +d}\, x\, dx \]

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 = 1.58, size = 143, normalized size = 1.01 \[ \frac {4}{225} \, {\left (\frac {15 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{2}} - \frac {9 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d - 30 \, \sqrt {e x + d} d^{2}}{e^{2}}\right )} b n + \frac {2}{15} \, b {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{2}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{2}}\right )} \log \left (c x^{n}\right ) + \frac {2}{15} \, a {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{2}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{2}}\right )} \]

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]

4/225*(15*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^2 - (9*(e*x + d)^(5/2) - 10*(e*x
+ d)^(3/2)*d - 30*sqrt(e*x + d)*d^2)/e^2)*b*n + 2/15*b*(3*(e*x + d)^(5/2)/e^2 - 5*(e*x + d)^(3/2)*d/e^2)*log(c
*x^n) + 2/15*a*(3*(e*x + d)^(5/2)/e^2 - 5*(e*x + d)^(3/2)*d/e^2)

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

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 [A]  time = 6.86, size = 224, normalized size = 1.58 \[ \frac {2 \left (- \frac {a d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {a \left (d + e x\right )^{\frac {5}{2}}}{5} - b d \left (\frac {\left (d + e x\right )^{\frac {3}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{3} - \frac {2 n \left (\frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d e \sqrt {d + e x} + \frac {e \left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{3 e}\right ) + b \left (\frac {\left (d + e x\right )^{\frac {5}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{5} - \frac {2 n \left (\frac {d^{3} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d^{2} e \sqrt {d + e x} + \frac {d e \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {e \left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{5 e}\right )\right )}{e^{2}} \]

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]

2*(-a*d*(d + e*x)**(3/2)/3 + a*(d + e*x)**(5/2)/5 - b*d*((d + e*x)**(3/2)*log(c*(-d/e + (d + e*x)/e)**n)/3 - 2
*n*(d**2*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d*e*sqrt(d + e*x) + e*(d + e*x)**(3/2)/3)/(3*e)) + b*((d +
e*x)**(5/2)*log(c*(-d/e + (d + e*x)/e)**n)/5 - 2*n*(d**3*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d**2*e*sqrt
(d + e*x) + d*e*(d + e*x)**(3/2)/3 + e*(d + e*x)**(5/2)/5)/(5*e)))/e**2

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