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

Optimal. Leaf size=194 \[ \frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{44 b d^2 n \sqrt{d+e x}}{5 e^4}+\frac{64 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{5 e^4}+\frac{16 b d n (d+e x)^{3/2}}{15 e^4}-\frac{4 b n (d+e x)^{5/2}}{25 e^4} \]

[Out]

(-44*b*d^2*n*Sqrt[d + e*x])/(5*e^4) + (16*b*d*n*(d + e*x)^(3/2))/(15*e^4) - (4*b*n*(d + e*x)^(5/2))/(25*e^4) +
(64*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(5*e^4) + (2*d^3*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x]) + (6
*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^4 - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/e^4 + (2*(d + e*x)^(5/2)
*(a + b*Log[c*x^n]))/(5*e^4)

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Rubi [A]  time = 0.195251, antiderivative size = 194, normalized size of antiderivative = 1., 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, 1620, 63, 208} \[ \frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{44 b d^2 n \sqrt{d+e x}}{5 e^4}+\frac{64 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{5 e^4}+\frac{16 b d n (d+e x)^{3/2}}{15 e^4}-\frac{4 b n (d+e x)^{5/2}}{25 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-44*b*d^2*n*Sqrt[d + e*x])/(5*e^4) + (16*b*d*n*(d + e*x)^(3/2))/(15*e^4) - (4*b*n*(d + e*x)^(5/2))/(25*e^4) +
(64*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(5*e^4) + (2*d^3*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x]) + (6
*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^4 - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/e^4 + (2*(d + e*x)^(5/2)
*(a + b*Log[c*x^n]))/(5*e^4)

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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 \frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx &=\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-(b n) \int \frac{2 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )}{5 e^4 x \sqrt{d+e x}} \, dx\\ &=\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{(2 b n) \int \frac{16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3}{x \sqrt{d+e x}} \, dx}{5 e^4}\\ &=\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{(2 b n) \int \left (\frac{11 d^2 e}{\sqrt{d+e x}}+\frac{16 d^3}{x \sqrt{d+e x}}-4 d e \sqrt{d+e x}+e (d+e x)^{3/2}\right ) \, dx}{5 e^4}\\ &=-\frac{44 b d^2 n \sqrt{d+e x}}{5 e^4}+\frac{16 b d n (d+e x)^{3/2}}{15 e^4}-\frac{4 b n (d+e x)^{5/2}}{25 e^4}+\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{\left (32 b d^3 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{5 e^4}\\ &=-\frac{44 b d^2 n \sqrt{d+e x}}{5 e^4}+\frac{16 b d n (d+e x)^{3/2}}{15 e^4}-\frac{4 b n (d+e x)^{5/2}}{25 e^4}+\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{\left (64 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 )}{5 e^5}\\ &=-\frac{44 b d^2 n \sqrt{d+e x}}{5 e^4}+\frac{16 b d n (d+e x)^{3/2}}{15 e^4}-\frac{4 b n (d+e x)^{5/2}}{25 e^4}+\frac{64 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{5 e^4}+\frac{2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x}}+\frac{6 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\\ \end{align*}

Mathematica [A]  time = 0.118449, size = 159, normalized size = 0.82 \[ \frac{240 a d^2 e x+480 a d^3-60 a d e^2 x^2+30 a e^3 x^3+30 b \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right ) \log \left (c x^n\right )-536 b d^2 e n x+960 b d^{5/2} n \sqrt{d+e x} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-592 b d^3 n+44 b d e^2 n x^2-12 b e^3 n x^3}{75 e^4 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*a*d^3 - 592*b*d^3*n + 240*a*d^2*e*x - 536*b*d^2*e*n*x - 60*a*d*e^2*x^2 + 44*b*d*e^2*n*x^2 + 30*a*e^3*x^3
- 12*b*e^3*n*x^3 + 960*b*d^(5/2)*n*Sqrt[d + e*x]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 30*b*(16*d^3 + 8*d^2*e*x - 2
*d*e^2*x^2 + e^3*x^3)*Log[c*x^n])/(75*e^4*Sqrt[d + e*x])

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Maple [F]  time = 0.502, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59731, size = 1031, normalized size = 5.31 \begin{align*} \left [\frac{2 \,{\left (240 \,{\left (b d^{2} e n x + b d^{3} n\right )} \sqrt{d} \log \left (\frac{e x + 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) -{\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \,{\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \,{\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \,{\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \,{\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \,{\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{75 \,{\left (e^{5} x + d e^{4}\right )}}, -\frac{2 \,{\left (480 \,{\left (b d^{2} e n x + b d^{3} n\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \,{\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \,{\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \,{\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \,{\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \,{\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{75 \,{\left (e^{5} x + d e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/75*(240*(b*d^2*e*n*x + b*d^3*n)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (296*b*d^3*n - 240*a
*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^3 - 2*(11*b*d*e^2*n - 15*a*d*e^2)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*
(b*e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*(b*e^3*n*x^3 - 2*b*d*e^2*n*x^2 + 8*b*d^2*e*n*
x + 16*b*d^3*n)*log(x))*sqrt(e*x + d))/(e^5*x + d*e^4), -2/75*(480*(b*d^2*e*n*x + b*d^3*n)*sqrt(-d)*arctan(sqr
t(e*x + d)*sqrt(-d)/d) + (296*b*d^3*n - 240*a*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^3 - 2*(11*b*d*e^2*n - 15*a*d*e^2
)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*(b*e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*
(b*e^3*n*x^3 - 2*b*d*e^2*n*x^2 + 8*b*d^2*e*n*x + 16*b*d^3*n)*log(x))*sqrt(e*x + d))/(e^5*x + d*e^4)]

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Sympy [A]  time = 66.9555, size = 386, normalized size = 1.99 \begin{align*} - \frac{- \frac{2 a d^{3}}{\sqrt{d + e x}} - 6 a d^{2} \sqrt{d + e x} + 2 a d \left (d + e x\right )^{\frac{3}{2}} - \frac{2 a \left (d + e x\right )^{\frac{5}{2}}}{5} + 2 b d^{3} \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 ) - 6 b d^{2} \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 ) + 6 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 ) - 2 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 )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2*a*d**3/sqrt(d + e*x) - 6*a*d**2*sqrt(d + e*x) + 2*a*d*(d + e*x)**(3/2) - 2*a*(d + e*x)**(5/2)/5 + 2*b*d**
3*(2*n*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) - log(c*(-d/e + (d + e*x)/e)**n)/sqrt(d + e*x)) - 6*b*d**2*(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)
+ 6*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)) - 2*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**4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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