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

Optimal. Leaf size=73 \[ \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}-\frac {8 b n (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{25 d}+\frac {16 b^2 n^2 (d x)^{5/2}}{125 d} \]

[Out]

16/125*b^2*n^2*(d*x)^(5/2)/d-8/25*b*n*(d*x)^(5/2)*(a+b*ln(c*x^n))/d+2/5*(d*x)^(5/2)*(a+b*ln(c*x^n))^2/d

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2305, 2304} \[ \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}-\frac {8 b n (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{25 d}+\frac {16 b^2 n^2 (d x)^{5/2}}{125 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*n^2*(d*x)^(5/2))/(125*d) - (8*b*n*(d*x)^(5/2)*(a + b*Log[c*x^n]))/(25*d) + (2*(d*x)^(5/2)*(a + b*Log[c
*x^n])^2)/(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
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {align*} \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx &=\frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}-\frac {1}{5} (4 b n) \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\\ &=\frac {16 b^2 n^2 (d x)^{5/2}}{125 d}-\frac {8 b n (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{25 d}+\frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.84 \[ \frac {2}{125} x (d x)^{3/2} \left (25 a^2+10 b (5 a-2 b n) \log \left (c x^n\right )-20 a b n+25 b^2 \log ^2\left (c x^n\right )+8 b^2 n^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(d*x)^(3/2)*(25*a^2 - 20*a*b*n + 8*b^2*n^2 + 10*b*(5*a - 2*b*n)*Log[c*x^n] + 25*b^2*Log[c*x^n]^2))/125

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fricas [A]  time = 0.47, size = 121, normalized size = 1.66 \[ \frac {2}{125} \, {\left (25 \, b^{2} d n^{2} x^{2} \log \relax (x)^{2} + 25 \, b^{2} d x^{2} \log \relax (c)^{2} - 10 \, {\left (2 \, b^{2} d n - 5 \, a b d\right )} x^{2} \log \relax (c) + {\left (8 \, b^{2} d n^{2} - 20 \, a b d n + 25 \, a^{2} d\right )} x^{2} + 10 \, {\left (5 \, b^{2} d n x^{2} \log \relax (c) - {\left (2 \, b^{2} d n^{2} - 5 \, a b d n\right )} x^{2}\right )} \log \relax (x)\right )} \sqrt {d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/125*(25*b^2*d*n^2*x^2*log(x)^2 + 25*b^2*d*x^2*log(c)^2 - 10*(2*b^2*d*n - 5*a*b*d)*x^2*log(c) + (8*b^2*d*n^2
- 20*a*b*d*n + 25*a^2*d)*x^2 + 10*(5*b^2*d*n*x^2*log(c) - (2*b^2*d*n^2 - 5*a*b*d*n)*x^2)*log(x))*sqrt(d*x)

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giac [C]  time = 1.42, size = 386, normalized size = 5.29 \[ -\frac {1}{125} \, {\left (-\left (25 i + 25\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) \log \relax (x)^{2} + \left (25 i - 25\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \log \relax (x)^{2} \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) + \left (20 i + 20\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) \log \relax (x) - \left (50 i + 50\right ) \, \sqrt {2} b^{2} n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) \log \relax (c) \log \relax (x) - \left (20 i - 20\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \log \relax (x) \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) + \left (50 i - 50\right ) \, \sqrt {2} b^{2} n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \log \relax (c) \log \relax (x) \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - \left (8 i + 8\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) + \left (20 i + 20\right ) \, \sqrt {2} b^{2} n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) \log \relax (c) - \left (50 i + 50\right ) \, \sqrt {2} a 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 (8 i - 8\right ) \, \sqrt {2} b^{2} n^{2} x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - \left (20 i - 20\right ) \, \sqrt {2} b^{2} n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \log \relax (c) \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) + \left (50 i - 50\right ) \, \sqrt {2} a 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 (20 i + 20\right ) \, \sqrt {2} a b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \cos \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - \left (20 i - 20\right ) \, \sqrt {2} a b n x^{\frac {5}{2}} \sqrt {{\left | d \right |}} \sin \left (\frac {1}{4} \, \pi \mathrm {sgn}\relax (d)\right ) - 50 \, b^{2} \sqrt {d} x^{\frac {5}{2}} \log \relax (c)^{2} - 100 \, a b \sqrt {d} x^{\frac {5}{2}} \log \relax (c) - 50 \, a^{2} \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))^2,x, algorithm="giac")

[Out]

-1/125*(-(25*I + 25)*sqrt(2)*b^2*n^2*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d))*log(x)^2 + (25*I - 25)*sqrt(2)*b^
2*n^2*x^(5/2)*sqrt(abs(d))*log(x)^2*sin(1/4*pi*sgn(d)) + (20*I + 20)*sqrt(2)*b^2*n^2*x^(5/2)*sqrt(abs(d))*cos(
1/4*pi*sgn(d))*log(x) - (50*I + 50)*sqrt(2)*b^2*n*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d))*log(c)*log(x) - (20*
I - 20)*sqrt(2)*b^2*n^2*x^(5/2)*sqrt(abs(d))*log(x)*sin(1/4*pi*sgn(d)) + (50*I - 50)*sqrt(2)*b^2*n*x^(5/2)*sqr
t(abs(d))*log(c)*log(x)*sin(1/4*pi*sgn(d)) - (8*I + 8)*sqrt(2)*b^2*n^2*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d))
+ (20*I + 20)*sqrt(2)*b^2*n*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d))*log(c) - (50*I + 50)*sqrt(2)*a*b*n*x^(5/2
)*sqrt(abs(d))*cos(1/4*pi*sgn(d))*log(x) + (8*I - 8)*sqrt(2)*b^2*n^2*x^(5/2)*sqrt(abs(d))*sin(1/4*pi*sgn(d)) -
(20*I - 20)*sqrt(2)*b^2*n*x^(5/2)*sqrt(abs(d))*log(c)*sin(1/4*pi*sgn(d)) + (50*I - 50)*sqrt(2)*a*b*n*x^(5/2)*
sqrt(abs(d))*log(x)*sin(1/4*pi*sgn(d)) + (20*I + 20)*sqrt(2)*a*b*n*x^(5/2)*sqrt(abs(d))*cos(1/4*pi*sgn(d)) - (
20*I - 20)*sqrt(2)*a*b*n*x^(5/2)*sqrt(abs(d))*sin(1/4*pi*sgn(d)) - 50*b^2*sqrt(d)*x^(5/2)*log(c)^2 - 100*a*b*s
qrt(d)*x^(5/2)*log(c) - 50*a^2*sqrt(d)*x^(5/2))*d

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maple [C]  time = 0.17, size = 716, normalized size = 9.81 \[ \frac {2 b^{2} d^{2} x^{3} \ln \left (x^{n}\right )^{2}}{5 \sqrt {d x}}+\frac {2 \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 ) b \,d^{2} x^{3} \ln \left (x^{n}\right )}{25 \sqrt {d x}}+\frac {\left (-25 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+50 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+50 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-100 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{4}+40 i \pi \,b^{2} n \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-100 i \pi \,b^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (c )-100 i \pi a b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+100 a^{2}-40 i \pi \,b^{2} n \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-40 i \pi \,b^{2} n \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+100 i \pi \,b^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (c )+100 i \pi \,b^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (c )+100 i \pi a b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+100 i \pi a b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+32 b^{2} n^{2}-25 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{4}+50 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{5}-25 \pi ^{2} b^{2} \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{4}+50 \pi ^{2} b^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{5}+200 a b \ln \relax (c )-80 b^{2} n \ln \relax (c )+100 b^{2} \ln \relax (c )^{2}-80 a b n -25 \pi ^{2} b^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{6}-100 i \pi \,b^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (c )-100 i \pi a b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+40 i \pi \,b^{2} n \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )\right ) d^{2} x^{3}}{250 \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)^2,x)

[Out]

2/5*d^2*b^2*x^3/(d*x)^(1/2)*ln(x^n)^2+2/25*d^2*b*x^3*(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)/(
d*x)^(1/2)*ln(x^n)+1/250*d^2*(-25*Pi^2*b^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2-100*Pi^2*b^2*csgn(I*c)*cs
gn(I*x^n)*csgn(I*c*x^n)^4+50*Pi^2*b^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+50*Pi^2*b^2*csgn(I*c)*csgn(I*x^n
)^2*csgn(I*c*x^n)^3+100*a^2+32*b^2*n^2-100*I*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+200*a*b*ln(c)-80
*b^2*n*ln(c)+100*b^2*ln(c)^2-25*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+50*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5
-80*a*b*n-25*Pi^2*b^2*csgn(I*c*x^n)^6+50*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5-25*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x
^n)^4-100*I*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+40*I*Pi*b^2*n*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-100*I
*ln(c)*Pi*b^2*csgn(I*c*x^n)^3-100*I*Pi*a*b*csgn(I*c*x^n)^3+40*I*Pi*b^2*n*csgn(I*c*x^n)^3+100*I*ln(c)*Pi*b^2*cs
gn(I*c*x^n)^2*csgn(I*c)+100*I*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2+100*I*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)^
2+100*I*Pi*a*b*csgn(I*c*x^n)^2*csgn(I*c)-40*I*Pi*b^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2-40*I*Pi*b^2*n*csgn(I*c*x^n)
^2*csgn(I*c))*x^3/(d*x)^(1/2)

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maxima [A]  time = 0.62, size = 102, normalized size = 1.40 \[ \frac {2 \, \left (d x\right )^{\frac {5}{2}} b^{2} \log \left (c x^{n}\right )^{2}}{5 \, d} - \frac {8 \, \left (d x\right )^{\frac {5}{2}} a b n}{25 \, d} + \frac {4 \, \left (d x\right )^{\frac {5}{2}} a b \log \left (c x^{n}\right )}{5 \, d} + \frac {2 \, \left (d x\right )^{\frac {5}{2}} a^{2}}{5 \, d} + \frac {8}{125} \, {\left (\frac {2 \, \left (d x\right )^{\frac {5}{2}} n^{2}}{d} - \frac {5 \, \left (d x\right )^{\frac {5}{2}} n \log \left (c x^{n}\right )}{d}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(d*x)^(5/2)*b^2*log(c*x^n)^2/d - 8/25*(d*x)^(5/2)*a*b*n/d + 4/5*(d*x)^(5/2)*a*b*log(c*x^n)/d + 2/5*(d*x)^(
5/2)*a^2/d + 8/125*(2*(d*x)^(5/2)*n^2/d - 5*(d*x)^(5/2)*n*log(c*x^n)/d)*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 50.36, size = 216, normalized size = 2.96 \[ \frac {2 a^{2} d^{\frac {3}{2}} x^{\frac {5}{2}}}{5} + \frac {4 a b d^{\frac {3}{2}} n x^{\frac {5}{2}} \log {\relax (x )}}{5} - \frac {8 a b d^{\frac {3}{2}} n x^{\frac {5}{2}}}{25} + \frac {4 a b d^{\frac {3}{2}} x^{\frac {5}{2}} \log {\relax (c )}}{5} + \frac {2 b^{2} d^{\frac {3}{2}} n^{2} x^{\frac {5}{2}} \log {\relax (x )}^{2}}{5} - \frac {8 b^{2} d^{\frac {3}{2}} n^{2} x^{\frac {5}{2}} \log {\relax (x )}}{25} + \frac {16 b^{2} d^{\frac {3}{2}} n^{2} x^{\frac {5}{2}}}{125} + \frac {4 b^{2} d^{\frac {3}{2}} n x^{\frac {5}{2}} \log {\relax (c )} \log {\relax (x )}}{5} - \frac {8 b^{2} d^{\frac {3}{2}} n x^{\frac {5}{2}} \log {\relax (c )}}{25} + \frac {2 b^{2} d^{\frac {3}{2}} x^{\frac {5}{2}} \log {\relax (c )}^{2}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*d**(3/2)*x**(5/2)/5 + 4*a*b*d**(3/2)*n*x**(5/2)*log(x)/5 - 8*a*b*d**(3/2)*n*x**(5/2)/25 + 4*a*b*d**(3/2
)*x**(5/2)*log(c)/5 + 2*b**2*d**(3/2)*n**2*x**(5/2)*log(x)**2/5 - 8*b**2*d**(3/2)*n**2*x**(5/2)*log(x)/25 + 16
*b**2*d**(3/2)*n**2*x**(5/2)/125 + 4*b**2*d**(3/2)*n*x**(5/2)*log(c)*log(x)/5 - 8*b**2*d**(3/2)*n*x**(5/2)*log
(c)/25 + 2*b**2*d**(3/2)*x**(5/2)*log(c)**2/5

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