3.10.27 \(\int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx\) [927]

3.10.27.1 Optimal result

Integrand size = 19, antiderivative size = 107 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}-\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}}+\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}} \]

output

-b^(1/4)*arctan(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/c^(3/2)+b^(1/ 
4)*arctanh(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/c^(3/2)-2*(b*x^2+a 
)^(1/4)/c/(c*x)^(1/2)
 

3.10.27.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\frac {x \left (-2 \sqrt [4]{a+b x^2}-\sqrt [4]{b} \sqrt {x} \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+\sqrt [4]{b} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{(c x)^{3/2}} \]

input

Integrate[(a + b*x^2)^(1/4)/(c*x)^(3/2),x]
 

output

(x*(-2*(a + b*x^2)^(1/4) - b^(1/4)*Sqrt[x]*ArcTan[(b^(1/4)*Sqrt[x])/(a + b 
*x^2)^(1/4)] + b^(1/4)*Sqrt[x]*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4) 
]))/(c*x)^(3/2)
 

3.10.27.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {247, 266, 854, 27, 827, 218, 221}

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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {b \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{3/4}}dx}{c^2}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 b \int \frac {c x}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{c^3}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {2 b \int \frac {c^3 x}{c^2-b c^2 x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{c^3}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b \int \frac {c x}{c^2-b c^2 x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{c}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {2 b \left (\frac {\int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\sqrt {b} x c+c}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}\right )}{c}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 b \left (\frac {\int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}\right )}{c}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}\right )}{c}-\frac {2 \sqrt [4]{a+b x^2}}{c \sqrt {c x}}\)

input

Int[(a + b*x^2)^(1/4)/(c*x)^(3/2),x]
 

output

(-2*(a + b*x^2)^(1/4))/(c*Sqrt[c*x]) + (2*b*(-1/2*ArcTan[(b^(1/4)*Sqrt[c*x 
])/(Sqrt[c]*(a + b*x^2)^(1/4))]/(b^(3/4)*Sqrt[c]) + ArcTanh[(b^(1/4)*Sqrt[ 
c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))]/(2*b^(3/4)*Sqrt[c])))/c
 

3.10.27.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

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 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

3.10.27.4 Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {3}{2}}}d x\]

input

int((b*x^2+a)^(1/4)/(c*x)^(3/2),x)
 

output

int((b*x^2+a)^(1/4)/(c*x)^(3/2),x)
 

3.10.27.5 Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\text {Timed out} \]

input

integrate((b*x^2+a)^(1/4)/(c*x)^(3/2),x, algorithm="fricas")
 

output

Timed out
 

3.10.27.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\frac {\sqrt [4]{a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

input

integrate((b*x**2+a)**(1/4)/(c*x)**(3/2),x)
 

output

a**(1/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a) 
/(2*c**(3/2)*sqrt(x)*gamma(3/4))
 

3.10.27.7 Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {3}{2}}} \,d x } \]

input

integrate((b*x^2+a)^(1/4)/(c*x)^(3/2),x, algorithm="maxima")
 

output

integrate((b*x^2 + a)^(1/4)/(c*x)^(3/2), x)
 

3.10.27.8 Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {3}{2}}} \,d x } \]

input

integrate((b*x^2+a)^(1/4)/(c*x)^(3/2),x, algorithm="giac")
 

output

integrate((b*x^2 + a)^(1/4)/(c*x)^(3/2), x)
 

3.10.27.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{3/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (c\,x\right )}^{3/2}} \,d x \]

input

int((a + b*x^2)^(1/4)/(c*x)^(3/2),x)
 

output

int((a + b*x^2)^(1/4)/(c*x)^(3/2), x)