\(\int (d+e x)^{3/2} (a+b \log (c x^n)) \, dx\) [140]

Optimal result

Integrand size = 20, antiderivative size = 115 \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {4 b d^2 n \sqrt {d+e x}}{5 e}-\frac {4 b d n (d+e x)^{3/2}}{15 e}-\frac {4 b n (d+e x)^{5/2}}{25 e}+\frac {4 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e} \]

[Out]

-4/15*b*d*n*(e*x+d)^(3/2)/e-4/25*b*n*(e*x+d)^(5/2)/e+4/5*b*d^(5/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e+2/5*(e*x
+d)^(5/2)*(a+b*ln(c*x^n))/e-4/5*b*d^2*n*(e*x+d)^(1/2)/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2356, 52, 65, 214} \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}+\frac {4 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e}-\frac {4 b d^2 n \sqrt {d+e x}}{5 e}-\frac {4 b d n (d+e x)^{3/2}}{15 e}-\frac {4 b n (d+e x)^{5/2}}{25 e} \]

[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 52

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2356

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]))

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 \left (-\frac {2}{15} b n \sqrt {d+e x} \left (23 d^2+11 d e x+3 e^2 x^2\right )+2 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+(d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )\right )}{5 e} \]

[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)

Maple [F]

\[\int \left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.50 \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (15 \, b d^{\frac {5}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (46 \, b d^{2} n - 15 \, a d^{2} + 3 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} + 2 \, {\left (11 \, b d e n - 15 \, a d e\right )} x - 15 \, {\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, e}, -\frac {2 \, {\left (30 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (46 \, b d^{2} n - 15 \, a d^{2} + 3 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} + 2 \, {\left (11 \, b d e n - 15 \, a d e\right )} x - 15 \, {\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, e}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 88.80 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.28 \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a d \left (\begin {cases} \frac {2 \left (d + e x\right )^{\frac {3}{2}}}{3 e} & \text {for}\: e \neq 0 \\\sqrt {d} x & \text {otherwise} \end {cases}\right ) + a e \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - b d n \left (\begin {cases} \frac {16 d^{\frac {3}{2}} \sqrt {1 + \frac {e x}{d}}}{9 e} + \frac {2 d^{\frac {3}{2}} \log {\left (\frac {e x}{d} \right )}}{3 e} - \frac {4 d^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{3 e} + \frac {4 \sqrt {d} x \sqrt {1 + \frac {e x}{d}}}{9} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\sqrt {d} x & \text {otherwise} \end {cases}\right ) + b d \left (\begin {cases} \frac {2 \left (d + e x\right )^{\frac {3}{2}}}{3 e} & \text {for}\: e \neq 0 \\\sqrt {d} x & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - b e n \left (\begin {cases} - \frac {124 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{225 e^{2}} - \frac {4 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{15 e^{2}} + \frac {8 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{15 e^{2}} + \frac {32 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{225 e} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{2}}{4} & \text {otherwise} \end {cases}\right ) + b e \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

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

[Out]

a*d*Piecewise((2*(d + e*x)**(3/2)/(3*e), Ne(e, 0)), (sqrt(d)*x, True)) + a*e*Piecewise((-2*d*(d + e*x)**(3/2)/
(3*e**2) + 2*(d + e*x)**(5/2)/(5*e**2), Ne(e, 0)), (sqrt(d)*x**2/2, True)) - b*d*n*Piecewise((16*d**(3/2)*sqrt
(1 + e*x/d)/(9*e) + 2*d**(3/2)*log(e*x/d)/(3*e) - 4*d**(3/2)*log(sqrt(1 + e*x/d) + 1)/(3*e) + 4*sqrt(d)*x*sqrt
(1 + e*x/d)/9, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x, True)) + b*d*Piecewise((2*(d + e*x)**(3/2)/(3*e),
Ne(e, 0)), (sqrt(d)*x, True))*log(c*x**n) - b*e*n*Piecewise((-124*d**(5/2)*sqrt(1 + e*x/d)/(225*e**2) - 4*d**
(5/2)*log(e*x/d)/(15*e**2) + 8*d**(5/2)*log(sqrt(1 + e*x/d) + 1)/(15*e**2) + 32*d**(3/2)*x*sqrt(1 + e*x/d)/(22
5*e) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/25, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**2/4, True)) + b*e*Piec
ewise((-2*d*(d + e*x)**(3/2)/(3*e**2) + 2*(d + e*x)**(5/2)/(5*e**2), Ne(e, 0)), (sqrt(d)*x**2/2, True))*log(c*
x**n)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 \, {\left (e x + d\right )}^{\frac {5}{2}} b \log \left (c x^{n}\right )}{5 \, e} + \frac {2 \, {\left (e x + d\right )}^{\frac {5}{2}} a}{5 \, e} - \frac {2 \, {\left (15 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right ) + 6 \, {\left (e x + d\right )}^{\frac {5}{2}} + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 30 \, \sqrt {e x + d} d^{2}\right )} b n}{75 \, e} \]

[In]

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

[Out]

2/5*(e*x + d)^(5/2)*b*log(c*x^n)/e + 2/5*(e*x + d)^(5/2)*a/e - 2/75*(15*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/
(sqrt(e*x + d) + sqrt(d))) + 6*(e*x + d)^(5/2) + 10*(e*x + d)^(3/2)*d + 30*sqrt(e*x + d)*d^2)*b*n/e

Giac [F]

\[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]

[In]

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

[Out]

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