\(\int \frac {(a+b \log (c x^n))^2}{(d x)^{3/2}} \, dx\) [99]

Optimal result

Integrand size = 20, antiderivative size = 67 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-\frac {16 b^2 n^2}{d \sqrt {d x}}-\frac {8 b n \left (a+b \log \left (c x^n\right )\right )}{d \sqrt {d x}}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{d \sqrt {d x}} \]

[Out]

-16*b^2*n^2/d/(d*x)^(1/2)-8*b*n*(a+b*ln(c*x^n))/d/(d*x)^(1/2)-2*(a+b*ln(c*x^n))^2/d/(d*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2342, 2341} \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-\frac {8 b n \left (a+b \log \left (c x^n\right )\right )}{d \sqrt {d x}}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{d \sqrt {d x}}-\frac {16 b^2 n^2}{d \sqrt {d x}} \]

[In]

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

[Out]

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

Rule 2341

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 2342

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 x \left (a^2+4 a b n+8 b^2 n^2+2 b (a+2 b n) \log \left (c x^n\right )+b^2 \log ^2\left (c x^n\right )\right )}{(d x)^{3/2}} \]

[In]

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

[Out]

(-2*x*(a^2 + 4*a*b*n + 8*b^2*n^2 + 2*b*(a + 2*b*n)*Log[c*x^n] + b^2*Log[c*x^n]^2))/(d*x)^(3/2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 707, normalized size of antiderivative = 10.55

method	result	size
risch	\(-\frac {2 b^{2} \ln \left (x^{n}\right )^{2}}{d \sqrt {d x}}-\frac {2 b \left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+4 b n +2 a \right ) \ln \left (x^{n}\right )}{d \sqrt {d x}}-\frac {4 a^{2}-8 i \pi \,b^{2} n \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \right )^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-4 i \ln \left (c \right ) \pi \,b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-4 i \pi a b \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-\pi ^{2} b^{2} \operatorname {csgn}\left (i c \right )^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-8 i \pi \,b^{2} n \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-4 i \ln \left (c \right ) \pi \,b^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-4 i \pi a b \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+2 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-4 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+2 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{5}-\pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+2 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{5}-\pi ^{2} b^{2} \operatorname {csgn}\left (i c \right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+32 b^{2} n^{2}+8 \ln \left (c \right ) a b +4 \ln \left (c \right )^{2} b^{2}+16 b^{2} \ln \left (c \right ) n +16 a b n -\pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{6}+4 i \pi a b \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+4 i \pi a b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi \,b^{2} n \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+4 i \ln \left (c \right ) \pi \,b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi \,b^{2} n \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+4 i \ln \left (c \right ) \pi \,b^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2 d \sqrt {d x}}\)	\(707\)

[In]

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

[Out]

-2/d*b^2/(d*x)^(1/2)*ln(x^n)^2-2/d*b*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^
n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+4*b*n+2*a)/(d*x)^(1/2)*ln(x^n)-1/2/d*
(4*a^2+2*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5+4*I*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)^2+4*I*Pi*a*b*csgn(I*x^n
)*csgn(I*c*x^n)^2+32*b^2*n^2+8*I*Pi*b^2*n*csgn(I*c)*csgn(I*c*x^n)^2+8*I*Pi*b^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2-4
*I*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+8*ln(c)*a*b+4*ln(c)^2*b^2+16*b^2*ln(c)*n+16*a*b*n-4*I*Pi*a
*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-4*I*ln(c)*Pi*b^2*csgn(I*c*x^n)^3-4*I*Pi*a*b*csgn(I*c*x^n)^3-Pi^2*b^2*cs
gn(I*x^n)^2*csgn(I*c*x^n)^4+2*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5+2*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c
*x^n)^3-4*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-8*I*Pi*b^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2*Pi
^2*b^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3-Pi^2*b^2*csgn(I*c*x^n)^6-Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x^n)^4+4
*I*Pi*a*b*csgn(I*c)*csgn(I*c*x^n)^2-Pi^2*b^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2-8*I*Pi*b^2*n*csgn(I*c*x
^n)^3+4*I*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*c*x^n)^2)/(d*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{2} n^{2} \log \left (x\right )^{2} + 8 \, b^{2} n^{2} + b^{2} \log \left (c\right )^{2} + 4 \, a b n + a^{2} + 2 \, {\left (2 \, b^{2} n + a b\right )} \log \left (c\right ) + 2 \, {\left (2 \, b^{2} n^{2} + b^{2} n \log \left (c\right ) + a b n\right )} \log \left (x\right )\right )} \sqrt {d x}}{d^{2} x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=- \frac {2 a^{2} x}{\left (d x\right )^{\frac {3}{2}}} - \frac {8 a b n x}{\left (d x\right )^{\frac {3}{2}}} - \frac {4 a b x \log {\left (c x^{n} \right )}}{\left (d x\right )^{\frac {3}{2}}} - \frac {16 b^{2} n^{2} x}{\left (d x\right )^{\frac {3}{2}}} - \frac {8 b^{2} n x \log {\left (c x^{n} \right )}}{\left (d x\right )^{\frac {3}{2}}} - \frac {2 b^{2} x \log {\left (c x^{n} \right )}^{2}}{\left (d x\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-8 \, b^{2} {\left (\frac {2 \, n^{2}}{\sqrt {d x} d} + \frac {n \log \left (c x^{n}\right )}{\sqrt {d x} d}\right )} - \frac {2 \, b^{2} \log \left (c x^{n}\right )^{2}}{\sqrt {d x} d} - \frac {8 \, a b n}{\sqrt {d x} d} - \frac {4 \, a b \log \left (c x^{n}\right )}{\sqrt {d x} d} - \frac {2 \, a^{2}}{\sqrt {d x} d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (61) = 122\).

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.22 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {b^{2} n^{2} \log \left (d x\right )^{2}}{\sqrt {d x}} - \frac {2 \, {\left (b^{2} n^{2} \log \left (d\right ) - 2 \, b^{2} n^{2} - b^{2} n \log \left (c\right ) - a b n\right )} \log \left (d x\right )}{\sqrt {d x}} + \frac {b^{2} n^{2} \log \left (d\right )^{2} - 4 \, b^{2} n^{2} \log \left (d\right ) - 2 \, b^{2} n \log \left (c\right ) \log \left (d\right ) + 8 \, b^{2} n^{2} + 4 \, b^{2} n \log \left (c\right ) + b^{2} \log \left (c\right )^{2} - 2 \, a b n \log \left (d\right ) + 4 \, a b n + 2 \, a b \log \left (c\right ) + a^{2}}{\sqrt {d x}}\right )}}{d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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