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

Optimal. Leaf size=146 \[ -\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}+\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3} \]

[Out]

-4/9*b*n*(e*x+d)^(3/2)/e^3-32/3*b*d^(3/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^3+2/3*(e*x+d)^(3/2)*(a+b*ln(c*x^n
))/e^3-2*d^2*(a+b*ln(c*x^n))/e^3/(e*x+d)^(1/2)+20/3*b*d*n*(e*x+d)^(1/2)/e^3-4*d*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/
e^3

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {43, 2350, 12, 897, 1153, 208} \[ -\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}+\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*b*d*n*Sqrt[d + e*x])/(3*e^3) - (4*b*n*(d + e*x)^(3/2))/(9*e^3) - (32*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqr
t[d]])/(3*e^3) - (2*d^2*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e*x]) - (4*d*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^3 +
(2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3)

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 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 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-(b n) \int \frac {2 \left (-8 d^2-4 d e x+e^2 x^2\right )}{3 e^3 x \sqrt {d+e x}} \, dx\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(2 b n) \int \frac {-8 d^2-4 d e x+e^2 x^2}{x \sqrt {d+e x}} \, dx}{3 e^3}\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(4 b n) \operatorname {Subst}\left (\int \frac {-3 d^2-6 d x^2+x^4}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 e^4}\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(4 b n) \operatorname {Subst}\left (\int \left (-5 d e+e x^2-\frac {8 d^2}{-\frac {d}{e}+\frac {x^2}{e}}\right ) \, dx,x,\sqrt {d+e x}\right )}{3 e^4}\\ &=\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (32 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^4}\\ &=\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 124, normalized size = 0.85 \[ \frac {-48 a d^2-24 a d e x+6 a e^2 x^2-6 b \left (8 d^2+4 d e x-e^2 x^2\right ) \log \left (c x^n\right )-96 b d^{3/2} n \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+56 b d^2 n+52 b d e n x-4 b e^2 n x^2}{9 e^3 \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-48*a*d^2 + 56*b*d^2*n - 24*a*d*e*x + 52*b*d*e*n*x + 6*a*e^2*x^2 - 4*b*e^2*n*x^2 - 96*b*d^(3/2)*n*Sqrt[d + e*
x]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] - 6*b*(8*d^2 + 4*d*e*x - e^2*x^2)*Log[c*x^n])/(9*e^3*Sqrt[d + e*x])

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 330, normalized size = 2.26 \[ \left [\frac {2 \, {\left (24 \, {\left (b d e n x + b d^{2} n\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (28 \, b d^{2} n - 24 \, a d^{2} - {\left (2 \, b e^{2} n - 3 \, a e^{2}\right )} x^{2} + 2 \, {\left (13 \, b d e n - 6 \, a d e\right )} x + 3 \, {\left (b e^{2} x^{2} - 4 \, b d e x - 8 \, b d^{2}\right )} \log \relax (c) + 3 \, {\left (b e^{2} n x^{2} - 4 \, b d e n x - 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, {\left (e^{4} x + d e^{3}\right )}}, \frac {2 \, {\left (48 \, {\left (b d e n x + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (28 \, b d^{2} n - 24 \, a d^{2} - {\left (2 \, b e^{2} n - 3 \, a e^{2}\right )} x^{2} + 2 \, {\left (13 \, b d e n - 6 \, a d e\right )} x + 3 \, {\left (b e^{2} x^{2} - 4 \, b d e x - 8 \, b d^{2}\right )} \log \relax (c) + 3 \, {\left (b e^{2} n x^{2} - 4 \, b d e n x - 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, {\left (e^{4} x + d e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/9*(24*(b*d*e*n*x + b*d^2*n)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (28*b*d^2*n - 24*a*d^2 -
(2*b*e^2*n - 3*a*e^2)*x^2 + 2*(13*b*d*e*n - 6*a*d*e)*x + 3*(b*e^2*x^2 - 4*b*d*e*x - 8*b*d^2)*log(c) + 3*(b*e^
2*n*x^2 - 4*b*d*e*n*x - 8*b*d^2*n)*log(x))*sqrt(e*x + d))/(e^4*x + d*e^3), 2/9*(48*(b*d*e*n*x + b*d^2*n)*sqrt(
-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (28*b*d^2*n - 24*a*d^2 - (2*b*e^2*n - 3*a*e^2)*x^2 + 2*(13*b*d*e*n - 6*
a*d*e)*x + 3*(b*e^2*x^2 - 4*b*d*e*x - 8*b*d^2)*log(c) + 3*(b*e^2*n*x^2 - 4*b*d*e*n*x - 8*b*d^2*n)*log(x))*sqrt
(e*x + d))/(e^4*x + d*e^3)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2}}{\left (e x +d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.45, size = 157, normalized size = 1.08 \[ \frac {4}{9} \, b n {\left (\frac {12 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} - \frac {{\left (e x + d\right )}^{\frac {3}{2}} - 15 \, \sqrt {e x + d} d}{e^{3}}\right )} + \frac {2}{3} \, b {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{3}} - \frac {6 \, \sqrt {e x + d} d}{e^{3}} - \frac {3 \, d^{2}}{\sqrt {e x + d} e^{3}}\right )} \log \left (c x^{n}\right ) + \frac {2}{3} \, a {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{3}} - \frac {6 \, \sqrt {e x + d} d}{e^{3}} - \frac {3 \, d^{2}}{\sqrt {e x + d} e^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*b*n*(12*d^(3/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^3 - ((e*x + d)^(3/2) - 15*sqrt(
e*x + d)*d)/e^3) + 2/3*b*((e*x + d)^(3/2)/e^3 - 6*sqrt(e*x + d)*d/e^3 - 3*d^2/(sqrt(e*x + d)*e^3))*log(c*x^n)
+ 2/3*a*((e*x + d)^(3/2)/e^3 - 6*sqrt(e*x + d)*d/e^3 - 3*d^2/(sqrt(e*x + d)*e^3))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 39.95, size = 262, normalized size = 1.79 \[ \frac {- \frac {2 a d^{2}}{\sqrt {d + e x}} - 4 a d \sqrt {d + e x} + \frac {2 a \left (d + e x\right )^{\frac {3}{2}}}{3} + 2 b d^{2} \left (\frac {2 n \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} - \frac {\log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{\sqrt {d + e x}}\right ) - 4 b d \left (\sqrt {d + e x} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )} - \frac {2 n \left (\frac {d e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + e \sqrt {d + e x}\right )}{e}\right ) + 2 b \left (\frac {\left (d + e x\right )^{\frac {3}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{3} - \frac {2 n \left (\frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d e \sqrt {d + e x} + \frac {e \left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{3 e}\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*a*d**2/sqrt(d + e*x) - 4*a*d*sqrt(d + e*x) + 2*a*(d + e*x)**(3/2)/3 + 2*b*d**2*(2*n*atan(sqrt(d + e*x)/sqr
t(-d))/sqrt(-d) - log(c*(-d/e + (d + e*x)/e)**n)/sqrt(d + e*x)) - 4*b*d*(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) + 2*b*((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)))/e**3

________________________________________________________________________________________