Integrand size = 20, antiderivative size = 73 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {16 b^2 n^2 (d x)^{5/2}}{125 d}-\frac {8 b n (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{25 d}+\frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d} \] Output:
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
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2}{125} x (d x)^{3/2} \left (25 a^2-20 a b n+8 b^2 n^2+10 b (5 a-2 b n) \log \left (c x^n\right )+25 b^2 \log ^2\left (c x^n\right )\right ) \] Input:
Integrate[(d*x)^(3/2)*(a + b*Log[c*x^n])^2,x]
Output:
(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
Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2742, 2741}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 2742 |
\(\displaystyle \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}-\frac {4}{5} b n \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )dx\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2}{5 d}-\frac {4}{5} b n \left (\frac {2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d}-\frac {4 b n (d x)^{5/2}}{25 d}\right )\) |
Input:
Int[(d*x)^(3/2)*(a + b*Log[c*x^n])^2,x]
Output:
(2*(d*x)^(5/2)*(a + b*Log[c*x^n])^2)/(5*d) - (4*b*n*((-4*b*n*(d*x)^(5/2))/ (25*d) + (2*(d*x)^(5/2)*(a + b*Log[c*x^n]))/(5*d)))/5
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]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbo l] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^p/(d*(m + 1))), x] - Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.28 (sec) , antiderivative size = 716, normalized size of antiderivative = 9.81
method | result | size |
risch | \(\frac {2 d^{2} b^{2} x^{3} \ln \left (x^{n}\right )^{2}}{5 \sqrt {d x}}+\frac {2 d^{2} b \,x^{3} \left (5 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-5 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-5 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+5 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+10 b \ln \left (c \right )-4 n b +10 a \right ) \ln \left (x^{n}\right )}{25 \sqrt {d x}}+\frac {d^{2} \left (-100 i \pi a b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-100 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+40 i \pi \,b^{2} n \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+100 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+100 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-80 b^{2} \ln \left (c \right ) n +200 a b \ln \left (c \right )+100 a^{2}+100 b^{2} \ln \left (c \right )^{2}+32 b^{2} n^{2}-25 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+50 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{5}-40 i \pi \,b^{2} n \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-40 i \pi \,b^{2} n \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+100 i \pi a b \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-80 a n b -100 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )-100 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-100 i \pi a b \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+50 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{5} \operatorname {csgn}\left (i c \right )-25 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )^{2}-25 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{6}+50 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )^{2}+50 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )+40 i \pi \,b^{2} n \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-25 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )^{2}+100 i \pi a b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}\right ) x^{3}}{250 \sqrt {d x}}\) | \(716\) |
Input:
int((d*x)^(3/2)*(a+b*ln(c*x^n))^2,x,method=_RETURNVERBOSE)
Output:
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*csg n(I*c*x^n)^3+5*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+10*b*ln(c)-4*n*b+10*a)/(d* x)^(1/2)*ln(x^n)+1/250*d^2*(50*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^3*csgn (I*c)-25*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2*csgn(I*c)^2-100*Pi^2*b^2*c sgn(I*x^n)*csgn(I*c*x^n)^4*csgn(I*c)-80*b^2*ln(c)*n+200*a*b*ln(c)+100*a^2+ 100*b^2*ln(c)^2+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*x^n)*csgn(I*c*x^n)^2+100*I*Pi*ln(c)*b^2*csgn(I*c*x^n)^2* csgn(I*c)+100*I*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2+32*b^2*n^2+50*Pi^2*b^2* csgn(I*c*x^n)^5*csgn(I*c)-25*Pi^2*b^2*csgn(I*c*x^n)^4*csgn(I*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- 100*I*Pi*a*b*csgn(I*c*x^n)^3+40*I*Pi*b^2*n*csgn(I*c*x^n)^3+50*Pi^2*b^2*csg n(I*x^n)*csgn(I*c*x^n)^3*csgn(I*c)^2-100*I*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I *c*x^n)*csgn(I*c)-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)^2-40*I*Pi*b^2*n*csgn(I*c*x^n)^2*csgn(I*c) -100*I*Pi*ln(c)*b^2*csgn(I*c*x^n)^3-80*a*n*b+100*I*Pi*a*b*csgn(I*c*x^n)^2* csgn(I*c)-25*Pi^2*b^2*csgn(I*c*x^n)^6)*x^3/(d*x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.66 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2}{125} \, {\left (25 \, b^{2} d n^{2} x^{2} \log \left (x\right )^{2} + 25 \, b^{2} d x^{2} \log \left (c\right )^{2} - 10 \, {\left (2 \, b^{2} d n - 5 \, a b d\right )} x^{2} \log \left (c\right ) + {\left (8 \, b^{2} d n^{2} - 20 \, a b d n + 25 \, a^{2} d\right )} x^{2} + 10 \, {\left (5 \, b^{2} d n x^{2} \log \left (c\right ) - {\left (2 \, b^{2} d n^{2} - 5 \, a b d n\right )} x^{2}\right )} \log \left (x\right )\right )} \sqrt {d x} \] Input:
integrate((d*x)^(3/2)*(a+b*log(c*x^n))^2,x, algorithm="fricas")
Output:
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)
Time = 3.91 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.63 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2 a^{2} x \left (d x\right )^{\frac {3}{2}}}{5} - \frac {8 a b n x \left (d x\right )^{\frac {3}{2}}}{25} + \frac {4 a b x \left (d x\right )^{\frac {3}{2}} \log {\left (c x^{n} \right )}}{5} + \frac {16 b^{2} n^{2} x \left (d x\right )^{\frac {3}{2}}}{125} - \frac {8 b^{2} n x \left (d x\right )^{\frac {3}{2}} \log {\left (c x^{n} \right )}}{25} + \frac {2 b^{2} x \left (d x\right )^{\frac {3}{2}} \log {\left (c x^{n} \right )}^{2}}{5} \] Input:
integrate((d*x)**(3/2)*(a+b*ln(c*x**n))**2,x)
Output:
2*a**2*x*(d*x)**(3/2)/5 - 8*a*b*n*x*(d*x)**(3/2)/25 + 4*a*b*x*(d*x)**(3/2) *log(c*x**n)/5 + 16*b**2*n**2*x*(d*x)**(3/2)/125 - 8*b**2*n*x*(d*x)**(3/2) *log(c*x**n)/25 + 2*b**2*x*(d*x)**(3/2)*log(c*x**n)**2/5
Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2 \, \left (d x\right )^{\frac {5}{2}} b^{2} \log \left (c x^{n}\right )^{2}}{5 \, d} - \frac {8 \, \left (d x\right )^{\frac {5}{2}} a b n}{25 \, d} + \frac {4 \, \left (d x\right )^{\frac {5}{2}} a b \log \left (c x^{n}\right )}{5 \, d} + \frac {2 \, \left (d x\right )^{\frac {5}{2}} a^{2}}{5 \, d} + \frac {8}{125} \, {\left (\frac {2 \, \left (d x\right )^{\frac {5}{2}} n^{2}}{d} - \frac {5 \, \left (d x\right )^{\frac {5}{2}} n \log \left (c x^{n}\right )}{d}\right )} b^{2} \] Input:
integrate((d*x)^(3/2)*(a+b*log(c*x^n))^2,x, algorithm="maxima")
Output:
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
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 386, normalized size of antiderivative = 5.29 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx =\text {Too large to display} \] Input:
integrate((d*x)^(3/2)*(a+b*log(c*x^n))^2,x, algorithm="giac")
Output:
-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*si n(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)*sqrt(ab s(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)*sqr t(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(ab s(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*sqrt(d)*x^( 5/2)*log(c) - 50*a^2*sqrt(d)*x^(5/2))*d
Timed out. \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\int {\left (d\,x\right )}^{3/2}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \] Input:
int((d*x)^(3/2)*(a + b*log(c*x^n))^2,x)
Output:
int((d*x)^(3/2)*(a + b*log(c*x^n))^2, x)
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2 \sqrt {x}\, \sqrt {d}\, d \,x^{2} \left (25 \mathrm {log}\left (x^{n} c \right )^{2} b^{2}+50 \,\mathrm {log}\left (x^{n} c \right ) a b -20 \,\mathrm {log}\left (x^{n} c \right ) b^{2} n +25 a^{2}-20 a b n +8 b^{2} n^{2}\right )}{125} \] Input:
int((d*x)^(3/2)*(a+b*log(c*x^n))^2,x)
Output:
(2*sqrt(x)*sqrt(d)*d*x**2*(25*log(x**n*c)**2*b**2 + 50*log(x**n*c)*a*b - 2 0*log(x**n*c)*b**2*n + 25*a**2 - 20*a*b*n + 8*b**2*n**2))/125