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

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

[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

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Rubi [A]
time = 0.04, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2356, 52, 65, 214} \begin {gather*} \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 \tanh ^{-1}\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} \end {gather*}

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 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*} \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 ) \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*}

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Mathematica [A]
time = 0.06, size = 87, normalized size = 0.76 \begin {gather*} \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 \tanh ^{-1}\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} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )\, dx\]

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 [A]
time = 0.50, size = 109, normalized size = 0.95 \begin {gather*} \frac {2}{5} \, {\left (x e + d\right )}^{\frac {5}{2}} b e^{\left (-1\right )} \log \left (c x^{n}\right ) + \frac {2}{5} \, {\left (x e + d\right )}^{\frac {5}{2}} a e^{\left (-1\right )} - \frac {2}{75} \, {\left (15 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) + 6 \, {\left (x e + d\right )}^{\frac {5}{2}} + 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 30 \, \sqrt {x e + d} d^{2}\right )} b n e^{\left (-1\right )} \end {gather*}

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]

2/5*(x*e + d)^(5/2)*b*e^(-1)*log(c*x^n) + 2/5*(x*e + d)^(5/2)*a*e^(-1) - 2/75*(15*d^(5/2)*log((sqrt(x*e + d) -
sqrt(d))/(sqrt(x*e + d) + sqrt(d))) + 6*(x*e + d)^(5/2) + 10*(x*e + d)^(3/2)*d + 30*sqrt(x*e + d)*d^2)*b*n*e^
(-1)

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Fricas [A]
time = 0.40, size = 283, normalized size = 2.46 \begin {gather*} \left [\frac {2}{75} \, {\left (15 \, b d^{\frac {5}{2}} n \log \left (\frac {x e + 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (46 \, b d^{2} n + 3 \, {\left (2 \, b n - 5 \, a\right )} x^{2} e^{2} - 15 \, a d^{2} + 2 \, {\left (11 \, b d n - 15 \, a d\right )} x e - 15 \, {\left (b x^{2} e^{2} + 2 \, b d x e + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{2} e^{2} + 2 \, b d n x e + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}, -\frac {2}{75} \, {\left (30 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (46 \, b d^{2} n + 3 \, {\left (2 \, b n - 5 \, a\right )} x^{2} e^{2} - 15 \, a d^{2} + 2 \, {\left (11 \, b d n - 15 \, a d\right )} x e - 15 \, {\left (b x^{2} e^{2} + 2 \, b d x e + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{2} e^{2} + 2 \, b d n x e + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}\right ] \end {gather*}

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((x*e + 2*sqrt(x*e + d)*sqrt(d) + 2*d)/x) - (46*b*d^2*n + 3*(2*b*n - 5*a)*x^2*e^2 - 1
5*a*d^2 + 2*(11*b*d*n - 15*a*d)*x*e - 15*(b*x^2*e^2 + 2*b*d*x*e + b*d^2)*log(c) - 15*(b*n*x^2*e^2 + 2*b*d*n*x*
e + b*d^2*n)*log(x))*sqrt(x*e + d))*e^(-1), -2/75*(30*b*sqrt(-d)*d^2*n*arctan(sqrt(x*e + d)*sqrt(-d)/d) + (46*
b*d^2*n + 3*(2*b*n - 5*a)*x^2*e^2 - 15*a*d^2 + 2*(11*b*d*n - 15*a*d)*x*e - 15*(b*x^2*e^2 + 2*b*d*x*e + b*d^2)*
log(c) - 15*(b*n*x^2*e^2 + 2*b*d*n*x*e + b*d^2*n)*log(x))*sqrt(x*e + d))*e^(-1)]

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Sympy [A]
time = 17.86, size = 333, normalized size = 2.90 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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