3.133 \(\int \sqrt {d+e x} (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=94 \[ \frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac {4 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e}-\frac {4 b d n \sqrt {d+e x}}{3 e}-\frac {4 b n (d+e x)^{3/2}}{9 e} \]

[Out]

-4/9*b*n*(e*x+d)^(3/2)/e+4/3*b*d^(3/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e+2/3*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e-
4/3*b*d*n*(e*x+d)^(1/2)/e

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2319, 50, 63, 208} \[ \frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac {4 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e}-\frac {4 b d n \sqrt {d+e x}}{3 e}-\frac {4 b n (d+e x)^{3/2}}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*d*n*Sqrt[d + e*x])/(3*e) - (4*b*n*(d + e*x)^(3/2))/(9*e) + (4*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]
)/(3*e) + (2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e)

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]

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]))

Rubi steps

\begin {align*} \int \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {(2 b n) \int \frac {(d+e x)^{3/2}}{x} \, dx}{3 e}\\ &=-\frac {4 b n (d+e x)^{3/2}}{9 e}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {(2 b d n) \int \frac {\sqrt {d+e x}}{x} \, dx}{3 e}\\ &=-\frac {4 b d n \sqrt {d+e x}}{3 e}-\frac {4 b n (d+e x)^{3/2}}{9 e}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {\left (2 b d^2 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{3 e}\\ &=-\frac {4 b d n \sqrt {d+e x}}{3 e}-\frac {4 b n (d+e x)^{3/2}}{9 e}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {\left (4 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 e^2}\\ &=-\frac {4 b d n \sqrt {d+e x}}{3 e}-\frac {4 b n (d+e x)^{3/2}}{9 e}+\frac {4 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 77, normalized size = 0.82 \[ \frac {2 \left (\sqrt {d+e x} \left (3 a (d+e x)+3 b (d+e x) \log \left (c x^n\right )-2 b n (4 d+e x)\right )+6 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(3*a*(d + e*x) - 2*b*n*(4*d + e*x) + 3*b*(d +
e*x)*Log[c*x^n])))/(9*e)

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fricas [A]  time = 0.58, size = 184, normalized size = 1.96 \[ \left [\frac {2 \, {\left (3 \, b d^{\frac {3}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (8 \, b d n - 3 \, a d + {\left (2 \, b e n - 3 \, a e\right )} x - 3 \, {\left (b e x + b d\right )} \log \relax (c) - 3 \, {\left (b e n x + b d n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, e}, -\frac {2 \, {\left (6 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (8 \, b d n - 3 \, a d + {\left (2 \, b e n - 3 \, a e\right )} x - 3 \, {\left (b e x + b d\right )} \log \relax (c) - 3 \, {\left (b e n x + b d n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{9 \, e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[2/9*(3*b*d^(3/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (8*b*d*n - 3*a*d + (2*b*e*n - 3*a*e)*x - 3*
(b*e*x + b*d)*log(c) - 3*(b*e*n*x + b*d*n)*log(x))*sqrt(e*x + d))/e, -2/9*(6*b*sqrt(-d)*d*n*arctan(sqrt(e*x +
d)*sqrt(-d)/d) + (8*b*d*n - 3*a*d + (2*b*e*n - 3*a*e)*x - 3*(b*e*x + b*d)*log(c) - 3*(b*e*n*x + b*d*n)*log(x))
*sqrt(e*x + d))/e]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(e*x+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*x + d)*(b*log(c*x^n) + a), x)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) \sqrt {e x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)*(e*x+d)^(1/2),x)

[Out]

int((b*ln(c*x^n)+a)*(e*x+d)^(1/2),x)

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maxima [A]  time = 1.40, size = 93, normalized size = 0.99 \[ \frac {2 \, {\left (e x + d\right )}^{\frac {3}{2}} b \log \left (c x^{n}\right )}{3 \, e} - \frac {2 \, {\left (3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right ) + 2 \, {\left (e x + d\right )}^{\frac {3}{2}} + 6 \, \sqrt {e x + d} d\right )} b n}{9 \, e} + \frac {2 \, {\left (e x + d\right )}^{\frac {3}{2}} a}{3 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/3*(e*x + d)^(3/2)*b*log(c*x^n)/e - 2/9*(3*d^(3/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d))) +
2*(e*x + d)^(3/2) + 6*sqrt(e*x + d)*d)*b*n/e + 2/3*(e*x + d)^(3/2)*a/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))*(d + e*x)^(1/2),x)

[Out]

int((a + b*log(c*x^n))*(d + e*x)^(1/2), x)

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sympy [A]  time = 3.79, size = 102, normalized size = 1.09 \[ \frac {2 \left (\frac {a \left (d + e x\right )^{\frac {3}{2}}}{3} + b \left (\frac {\left (d + e x\right )^{\frac {3}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{3} - \frac {2 n \left (\frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d e \sqrt {d + e x} + \frac {e \left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{3 e}\right )\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*(e*x+d)**(1/2),x)

[Out]

2*(a*(d + e*x)**(3/2)/3 + b*((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

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