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

Optimal. Leaf size=177 \[ -\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{35 e^2}+\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2} \]

[Out]

2/105*b*d^2*n*(e*x^2+d)^(3/2)/e^2+2/175*b*d*n*(e*x^2+d)^(5/2)/e^2-1/49*b*n*(e*x^2+d)^(7/2)/e^2-2/35*b*d^(7/2)*
n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^2-1/5*d*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^2+1/7*(e*x^2+d)^(7/2)*(a+b*ln(c
*x^n))/e^2+2/35*b*d^3*n*(e*x^2+d)^(1/2)/e^2

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Rubi [A]  time = 0.20, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 43, 2350, 12, 446, 80, 50, 63, 208} \[ -\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}-\frac {2 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{35 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d^3*n*Sqrt[d + e*x^2])/(35*e^2) + (2*b*d^2*n*(d + e*x^2)^(3/2))/(105*e^2) + (2*b*d*n*(d + e*x^2)^(5/2))/(
175*e^2) - (b*n*(d + e*x^2)^(7/2))/(49*e^2) - (2*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(35*e^2) - (d*(
d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2) + ((d + e*x^2)^(7/2)*(a + b*Log[c*x^n]))/(7*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-(b n) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{35 e^2 x} \, dx\\ &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {(b n) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{x} \, dx}{35 e^2}\\ &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {(b n) \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2} (-2 d+5 e x)}{x} \, dx,x,x^2\right )}{70 e^2}\\ &=-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {(b d n) \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2}}{x} \, dx,x,x^2\right )}{35 e^2}\\ &=\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (b d^2 n\right ) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{35 e^2}\\ &=\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (b d^3 n\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )}{35 e^2}\\ &=\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (b d^4 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{35 e^2}\\ &=\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (2 b d^4 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{35 e^3}\\ &=\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {2 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{35 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 227, normalized size = 1.28 \[ \sqrt {d+e x^2} \left (-\frac {d^3 \left (210 a+210 b \left (\log \left (c x^n\right )-n \log (x)\right )-247 b n\right )}{3675 e^2}+\frac {d^2 x^2 \left (105 a+105 b \left (\log \left (c x^n\right )-n \log (x)\right )-71 b n\right )}{3675 e}+\frac {d x^4 \left (280 a+280 b \left (\log \left (c x^n\right )-n \log (x)\right )-61 b n\right )}{1225}+\frac {1}{49} e x^6 \left (7 a+7 b \left (\log \left (c x^n\right )-n \log (x)\right )-b n\right )\right )-\frac {2 b d^{7/2} n \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )}{35 e^2}+\frac {2 b d^{7/2} n \log (x)}{35 e^2}-\frac {b n \log (x) \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^{5/2}}{35 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d^(7/2)*n*Log[x])/(35*e^2) - (b*n*(2*d - 5*e*x^2)*(d + e*x^2)^(5/2)*Log[x])/(35*e^2) + Sqrt[d + e*x^2]*((
e*x^6*(7*a - b*n + 7*b*(-(n*Log[x]) + Log[c*x^n])))/49 + (d^2*x^2*(105*a - 71*b*n + 105*b*(-(n*Log[x]) + Log[c
*x^n])))/(3675*e) - (d^3*(210*a - 247*b*n + 210*b*(-(n*Log[x]) + Log[c*x^n])))/(3675*e^2) + (d*x^4*(280*a - 61
*b*n + 280*b*(-(n*Log[x]) + Log[c*x^n])))/1225) - (2*b*d^(7/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(35*e^2)

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fricas [A]  time = 0.50, size = 409, normalized size = 2.31 \[ \left [\frac {105 \, b d^{\frac {7}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (75 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} - 247 \, b d^{3} n + 3 \, {\left (61 \, b d e^{2} n - 280 \, a d e^{2}\right )} x^{4} + 210 \, a d^{3} + {\left (71 \, b d^{2} e n - 105 \, a d^{2} e\right )} x^{2} - 105 \, {\left (5 \, b e^{3} x^{6} + 8 \, b d e^{2} x^{4} + b d^{2} e x^{2} - 2 \, b d^{3}\right )} \log \relax (c) - 105 \, {\left (5 \, b e^{3} n x^{6} + 8 \, b d e^{2} n x^{4} + b d^{2} e n x^{2} - 2 \, b d^{3} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{3675 \, e^{2}}, \frac {210 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left (75 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} - 247 \, b d^{3} n + 3 \, {\left (61 \, b d e^{2} n - 280 \, a d e^{2}\right )} x^{4} + 210 \, a d^{3} + {\left (71 \, b d^{2} e n - 105 \, a d^{2} e\right )} x^{2} - 105 \, {\left (5 \, b e^{3} x^{6} + 8 \, b d e^{2} x^{4} + b d^{2} e x^{2} - 2 \, b d^{3}\right )} \log \relax (c) - 105 \, {\left (5 \, b e^{3} n x^{6} + 8 \, b d e^{2} n x^{4} + b d^{2} e n x^{2} - 2 \, b d^{3} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{3675 \, e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3675*(105*b*d^(7/2)*n*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (75*(b*e^3*n - 7*a*e^3)*x^6 - 2
47*b*d^3*n + 3*(61*b*d*e^2*n - 280*a*d*e^2)*x^4 + 210*a*d^3 + (71*b*d^2*e*n - 105*a*d^2*e)*x^2 - 105*(5*b*e^3*
x^6 + 8*b*d*e^2*x^4 + b*d^2*e*x^2 - 2*b*d^3)*log(c) - 105*(5*b*e^3*n*x^6 + 8*b*d*e^2*n*x^4 + b*d^2*e*n*x^2 - 2
*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/3675*(210*b*sqrt(-d)*d^3*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - (75*(b
*e^3*n - 7*a*e^3)*x^6 - 247*b*d^3*n + 3*(61*b*d*e^2*n - 280*a*d*e^2)*x^4 + 210*a*d^3 + (71*b*d^2*e*n - 105*a*d
^2*e)*x^2 - 105*(5*b*e^3*x^6 + 8*b*d*e^2*x^4 + b*d^2*e*x^2 - 2*b*d^3)*log(c) - 105*(5*b*e^3*n*x^6 + 8*b*d*e^2*
n*x^4 + b*d^2*e*n*x^2 - 2*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/e^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (b \ln \left (c \,x^{n}\right )+a \right ) x^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 181, normalized size = 1.02 \[ \frac {1}{3675} \, {\left (\frac {105 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{2}} - \frac {75 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d - 70 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} - 210 \, \sqrt {e x^{2} + d} d^{3}}{e^{2}}\right )} b n + \frac {1}{35} \, {\left (\frac {5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{e} - \frac {2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{35} \, {\left (\frac {5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{e} - \frac {2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3675*(105*d^(7/2)*log((sqrt(e*x^2 + d) - sqrt(d))/(sqrt(e*x^2 + d) + sqrt(d)))/e^2 - (75*(e*x^2 + d)^(7/2) -
42*(e*x^2 + d)^(5/2)*d - 70*(e*x^2 + d)^(3/2)*d^2 - 210*sqrt(e*x^2 + d)*d^3)/e^2)*b*n + 1/35*(5*(e*x^2 + d)^(
5/2)*x^2/e - 2*(e*x^2 + d)^(5/2)*d/e^2)*b*log(c*x^n) + 1/35*(5*(e*x^2 + d)^(5/2)*x^2/e - 2*(e*x^2 + d)^(5/2)*d
/e^2)*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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