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

Optimal. Leaf size=163 \[ -\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{8 b d^3 n \sqrt{d+e x}}{35 e^2}+\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}-\frac{8 b d^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{35 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2} \]

[Out]

(8*b*d^3*n*Sqrt[d + e*x])/(35*e^2) + (8*b*d^2*n*(d + e*x)^(3/2))/(105*e^2) + (8*b*d*n*(d + e*x)^(5/2))/(175*e^
2) - (4*b*n*(d + e*x)^(7/2))/(49*e^2) - (8*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(35*e^2) - (2*d*(d + e*
x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2) + (2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^2)

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Rubi [A]  time = 0.117007, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{8 b d^3 n \sqrt{d+e x}}{35 e^2}+\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}-\frac{8 b d^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{35 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^3*n*Sqrt[d + e*x])/(35*e^2) + (8*b*d^2*n*(d + e*x)^(3/2))/(105*e^2) + (8*b*d*n*(d + e*x)^(5/2))/(175*e^
2) - (4*b*n*(d + e*x)^(7/2))/(49*e^2) - (8*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(35*e^2) - (2*d*(d + e*
x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2) + (2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^2)

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 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])

Rule 12

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

Rubi steps

\begin{align*} \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-(b n) \int \frac{2 (d+e x)^{5/2} (-2 d+5 e x)}{35 e^2 x} \, dx\\ &=-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac{(2 b n) \int \frac{(d+e x)^{5/2} (-2 d+5 e x)}{x} \, dx}{35 e^2}\\ &=-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{(4 b d n) \int \frac{(d+e x)^{5/2}}{x} \, dx}{35 e^2}\\ &=\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{\left (4 b d^2 n\right ) \int \frac{(d+e x)^{3/2}}{x} \, dx}{35 e^2}\\ &=\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{\left (4 b d^3 n\right ) \int \frac{\sqrt{d+e x}}{x} \, dx}{35 e^2}\\ &=\frac{8 b d^3 n \sqrt{d+e x}}{35 e^2}+\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{\left (4 b d^4 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{35 e^2}\\ &=\frac{8 b d^3 n \sqrt{d+e x}}{35 e^2}+\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac{\left (8 b d^4 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{35 e^3}\\ &=\frac{8 b d^3 n \sqrt{d+e x}}{35 e^2}+\frac{8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac{8 b d n (d+e x)^{5/2}}{175 e^2}-\frac{4 b n (d+e x)^{7/2}}{49 e^2}-\frac{8 b d^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{35 e^2}-\frac{2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}\\ \end{align*}

Mathematica [A]  time = 0.167948, size = 120, normalized size = 0.74 \[ -\frac{2 \left (\sqrt{d+e x} \left (105 a (2 d-5 e x) (d+e x)^2+105 b (2 d-5 e x) (d+e x)^2 \log \left (c x^n\right )+2 b n \left (71 d^2 e x-247 d^3+183 d e^2 x^2+75 e^3 x^3\right )\right )+420 b d^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )}{3675 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(420*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(105*a*(2*d - 5*e*x)*(d + e*x)^2 + 2*b*n*(
-247*d^3 + 71*d^2*e*x + 183*d*e^2*x^2 + 75*e^3*x^3) + 105*b*(2*d - 5*e*x)*(d + e*x)^2*Log[c*x^n])))/(3675*e^2)

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Maple [F]  time = 0.557, size = 0, normalized size = 0. \begin{align*} \int x \left ( ex+d \right ) ^{{\frac{3}{2}}} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5215, size = 960, normalized size = 5.89 \begin{align*} \left [\frac{2 \,{\left (210 \, b d^{\frac{7}{2}} n \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) +{\left (494 \, b d^{3} n - 210 \, a d^{3} - 75 \,{\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} - 6 \,{\left (61 \, b d e^{2} n - 140 \, a d e^{2}\right )} x^{2} -{\left (142 \, b d^{2} e n - 105 \, a d^{2} e\right )} x + 105 \,{\left (5 \, b e^{3} x^{3} + 8 \, b d e^{2} x^{2} + b d^{2} e x - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \,{\left (5 \, b e^{3} n x^{3} + 8 \, b d e^{2} n x^{2} + b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{3675 \, e^{2}}, \frac{2 \,{\left (420 \, b \sqrt{-d} d^{3} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (494 \, b d^{3} n - 210 \, a d^{3} - 75 \,{\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} - 6 \,{\left (61 \, b d e^{2} n - 140 \, a d e^{2}\right )} x^{2} -{\left (142 \, b d^{2} e n - 105 \, a d^{2} e\right )} x + 105 \,{\left (5 \, b e^{3} x^{3} + 8 \, b d e^{2} x^{2} + b d^{2} e x - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \,{\left (5 \, b e^{3} n x^{3} + 8 \, b d e^{2} n x^{2} + b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{3675 \, e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3675*(210*b*d^(7/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (494*b*d^3*n - 210*a*d^3 - 75*(2*b*e^3
*n - 7*a*e^3)*x^3 - 6*(61*b*d*e^2*n - 140*a*d*e^2)*x^2 - (142*b*d^2*e*n - 105*a*d^2*e)*x + 105*(5*b*e^3*x^3 +
8*b*d*e^2*x^2 + b*d^2*e*x - 2*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 + 8*b*d*e^2*n*x^2 + b*d^2*e*n*x - 2*b*d^3*n)*
log(x))*sqrt(e*x + d))/e^2, 2/3675*(420*b*sqrt(-d)*d^3*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (494*b*d^3*n - 210
*a*d^3 - 75*(2*b*e^3*n - 7*a*e^3)*x^3 - 6*(61*b*d*e^2*n - 140*a*d*e^2)*x^2 - (142*b*d^2*e*n - 105*a*d^2*e)*x +
105*(5*b*e^3*x^3 + 8*b*d*e^2*x^2 + b*d^2*e*x - 2*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 + 8*b*d*e^2*n*x^2 + b*d^2
*e*n*x - 2*b*d^3*n)*log(x))*sqrt(e*x + d))/e^2]

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Sympy [B]  time = 121.169, size = 583, normalized size = 3.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*d*(-d*(d + e*x)**(3/2)/3 + (d + e*x)**(5/2)/5)/e**2 + 2*a*(d**2*(d + e*x)**(3/2)/3 - 2*d*(d + e*x)**(5/2)/
5 + (d + e*x)**(7/2)/7)/e**2 + 2*b*d*(-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)) + (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 + 2*b*(d**2*((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)) -
2*d*((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)) + (d + e*x)**(7/2)*log(c*(-d/e +
(d + e*x)/e)**n)/7 - 2*n*(d**4*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d**3*e*sqrt(d + e*x) + d**2*e*(d + e*
x)**(3/2)/3 + d*e*(d + e*x)**(5/2)/5 + e*(d + e*x)**(7/2)/7)/(7*e))/e**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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