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

Optimal. Leaf size=41 \[ \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d}-\frac {4 b n (d x)^{5/2}}{25 d} \]

[Out]

-4/25*b*n*(d*x)^(5/2)/d+2/5*(d*x)^(5/2)*(a+b*ln(c*x^n))/d

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2304} \[ \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d}-\frac {4 b n (d x)^{5/2}}{25 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.71 \[ \frac {2}{25} x (d x)^{3/2} \left (5 a+5 b \log \left (c x^n\right )-2 b n\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 42, normalized size = 1.02 \[ \frac {2}{25} \, {\left (5 \, b d n x^{2} \log \relax (x) + 5 \, b d x^{2} \log \relax (c) - {\left (2 \, b d n - 5 \, a d\right )} x^{2}\right )} \sqrt {d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(5*b*d*n*x^2*log(x) + 5*b*d*x^2*log(c) - (2*b*d*n - 5*a*d)*x^2)*sqrt(d*x)

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giac [C]  time = 0.53, size = 108, normalized size = 2.63 \[ -\frac {1}{25} \, {\left (-\left (5 i + 5\right ) \, \sqrt {2} b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) \log \relax (x) + \left (5 i - 5\right ) \, \sqrt {2} b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \log \relax (x) \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) + \left (2 i + 2\right ) \, \sqrt {2} b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - \left (2 i - 2\right ) \, \sqrt {2} b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - 10 \, b \sqrt {d} x^{\frac {5}{2}} \log \relax (c) - 10 \, a \sqrt {d} x^{\frac {5}{2}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(-(5*I + 5)*sqrt(2)*b*n*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d))*log(x) + (5*I - 5)*sqrt(2)*b*n*x^(5/2)*s
qrt(abs(d))*log(x)*sin(1/4*pi*sgn(d)) + (2*I + 2)*sqrt(2)*b*n*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d)) - (2*I -
2)*sqrt(2)*b*n*x^(5/2)*sqrt(abs(d))*sin(1/4*pi*sgn(d)) - 10*b*sqrt(d)*x^(5/2)*log(c) - 10*a*sqrt(d)*x^(5/2))*
d

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maple [C]  time = 0.12, size = 128, normalized size = 3.12 \[ \frac {2 b \,d^{2} x^{3} \ln \left (x^{n}\right )}{5 \sqrt {d x}}+\frac {\left (-5 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+5 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+5 i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-5 i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-4 b n +10 b \ln \relax (c )+10 a \right ) d^{2} x^{3}}{25 \sqrt {d x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*d^2*b*x^3/(d*x)^(1/2)*ln(x^n)+1/25*d^2*(5*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-5*I*b*Pi*csgn(I*x^n)*csgn(I*c
*x^n)*csgn(I*c)-5*I*b*Pi*csgn(I*c*x^n)^3+5*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+10*b*ln(c)-4*b*n+10*a)*x^3/(d*x)^(
1/2)

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maxima [A]  time = 0.56, size = 41, normalized size = 1.00 \[ -\frac {4 \, \left (d x\right )^{\frac {5}{2}} b n}{25 \, d} + \frac {2 \, \left (d x\right )^{\frac {5}{2}} b \log \left (c x^{n}\right )}{5 \, d} + \frac {2 \, \left (d x\right )^{\frac {5}{2}} a}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/25*(d*x)^(5/2)*b*n/d + 2/5*(d*x)^(5/2)*b*log(c*x^n)/d + 2/5*(d*x)^(5/2)*a/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (d\,x\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*(a + b*log(c*x^n)),x)

[Out]

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

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sympy [A]  time = 25.45, size = 70, normalized size = 1.71 \[ \frac {2 a d^{\frac {3}{2}} x^{\frac {5}{2}}}{5} + \frac {2 b d^{\frac {3}{2}} n x^{\frac {5}{2}} \log {\relax (x )}}{5} - \frac {4 b d^{\frac {3}{2}} n x^{\frac {5}{2}}}{25} + \frac {2 b d^{\frac {3}{2}} x^{\frac {5}{2}} \log {\relax (c )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*d**(3/2)*x**(5/2)/5 + 2*b*d**(3/2)*n*x**(5/2)*log(x)/5 - 4*b*d**(3/2)*n*x**(5/2)/25 + 2*b*d**(3/2)*x**(5/2
)*log(c)/5

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