\(\int \frac {(c x)^{5/2}}{(a+b x^2)^{3/4}} \, dx\) [1039]

Optimal result

Integrand size = 19, antiderivative size = 117 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}+\frac {3 a c^{5/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}-\frac {3 a c^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}} \] Output:

1/2*c*(c*x)^(3/2)*(b*x^2+a)^(1/4)/b+3/4*a*c^(5/2)*arctan(b^(1/4)*(c*x)^(1/ 
2)/c^(1/2)/(b*x^2+a)^(1/4))/b^(7/4)-3/4*a*c^(5/2)*arctanh(b^(1/4)*(c*x)^(1 
/2)/c^(1/2)/(b*x^2+a)^(1/4))/b^(7/4)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.83 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {(c x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2}+3 a \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )-3 a \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{7/4} x^{5/2}} \] Input:

Integrate[(c*x)^(5/2)/(a + b*x^2)^(3/4),x]
 

Output:

((c*x)^(5/2)*(2*b^(3/4)*x^(3/2)*(a + b*x^2)^(1/4) + 3*a*ArcTan[(b^(1/4)*Sq 
rt[x])/(a + b*x^2)^(1/4)] - 3*a*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4 
)]))/(4*b^(7/4)*x^(5/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {262, 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.

\(\displaystyle \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(c*x)^(5/2)/(a + b*x^2)^(3/4),x]
 

Output:

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

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 262

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

Maple [F]

\[\int \frac {\left (c x \right )^{\frac {5}{2}}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]

Input:

int((c*x)^(5/2)/(b*x^2+a)^(3/4),x)
 

Output:

int((c*x)^(5/2)/(b*x^2+a)^(3/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \] Input:

integrate((c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.38 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate((c*x)**(5/2)/(b*x**2+a)**(3/4),x)
 

Output:

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(3/4)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((c*x)^(5/2)/(b*x^2 + a)^(3/4), x)
 

Giac [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((c*x)^(5/2)/(b*x^2 + a)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {{\left (c\,x\right )}^{5/2}}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:

int((c*x)^(5/2)/(a + b*x^2)^(3/4),x)
 

Output:

int((c*x)^(5/2)/(a + b*x^2)^(3/4), x)
 

Reduce [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx=\sqrt {c}\, \left (\int \frac {\sqrt {x}\, x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c^{2} \] Input:

int((c*x)^(5/2)/(b*x^2+a)^(3/4),x)
 

Output:

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