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

Optimal. Leaf size=115 \[ \frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{4 b d^2 n \sqrt{d+e x}}{5 e}+\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{5 e}-\frac{4 b d n (d+e x)^{3/2}}{15 e}-\frac{4 b n (d+e x)^{5/2}}{25 e} \]

[Out]

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

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Rubi [A]  time = 0.050536, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2319, 50, 63, 208} \[ \frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{4 b d^2 n \sqrt{d+e x}}{5 e}+\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{5 e}-\frac{4 b d n (d+e x)^{3/2}}{15 e}-\frac{4 b n (d+e x)^{5/2}}{25 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

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

Mathematica [A]  time = 0.0828026, size = 87, normalized size = 0.76 \[ \frac{2 \left ((d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )-\frac{2}{15} b n \sqrt{d+e x} \left (23 d^2+11 d e x+3 e^2 x^2\right )+2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-2*b*n*Sqrt[d + e*x]*(23*d^2 + 11*d*e*x + 3*e^2*x^2))/15 + 2*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] +
(d + e*x)^(5/2)*(a + b*Log[c*x^n])))/(5*e)

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Maple [F]  time = 0.563, size = 0, normalized size = 0. \begin{align*} \int \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((e*x+d)^(3/2)*(a+b*ln(c*x^n)),x)

[Out]

int((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((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 = 1.41295, size = 710, normalized size = 6.17 \begin{align*} \left [\frac{2 \,{\left (15 \, b d^{\frac{5}{2}} n \log \left (\frac{e x + 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) -{\left (46 \, b d^{2} n - 15 \, a d^{2} + 3 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} + 2 \,{\left (11 \, b d e n - 15 \, a d e\right )} x - 15 \,{\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{75 \, e}, -\frac{2 \,{\left (30 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (46 \, b d^{2} n - 15 \, a d^{2} + 3 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} + 2 \,{\left (11 \, b d e n - 15 \, a d e\right )} x - 15 \,{\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{75 \, e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/75*(15*b*d^(5/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (46*b*d^2*n - 15*a*d^2 + 3*(2*b*e^2*n - 5
*a*e^2)*x^2 + 2*(11*b*d*e*n - 15*a*d*e)*x - 15*(b*e^2*x^2 + 2*b*d*e*x + b*d^2)*log(c) - 15*(b*e^2*n*x^2 + 2*b*
d*e*n*x + b*d^2*n)*log(x))*sqrt(e*x + d))/e, -2/75*(30*b*sqrt(-d)*d^2*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (46
*b*d^2*n - 15*a*d^2 + 3*(2*b*e^2*n - 5*a*e^2)*x^2 + 2*(11*b*d*e*n - 15*a*d*e)*x - 15*(b*e^2*x^2 + 2*b*d*e*x +
b*d^2)*log(c) - 15*(b*e^2*n*x^2 + 2*b*d*e*n*x + b*d^2*n)*log(x))*sqrt(e*x + d))/e]

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Sympy [A]  time = 51.687, size = 333, normalized size = 2.9 \begin{align*} a d \left (\begin{cases} \sqrt{d} x & \text{for}\: e = 0 \\\frac{2 \left (d + e x\right )^{\frac{3}{2}}}{3 e} & \text{otherwise} \end{cases}\right ) + \frac{2 a \left (- \frac{d \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{\left (d + e x\right )^{\frac{5}{2}}}{5}\right )}{e} + \frac{2 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 )}{e} + \frac{2 b \left (- 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 ) + \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d*Piecewise((sqrt(d)*x, Eq(e, 0)), (2*(d + e*x)**(3/2)/(3*e), True)) + 2*a*(-d*(d + e*x)**(3/2)/3 + (d + e*x
)**(5/2)/5)/e + 2*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))/e + 2*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)) + (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))/sqr
t(-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

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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 )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)