\(\int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx\) [1015]

Optimal result

Integrand size = 19, antiderivative size = 147 \[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\frac {a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}+\frac {3 a^2 c^{5/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}-\frac {3 a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}} \] Output:

1/16*a*c*(c*x)^(3/2)*(b*x^2+a)^(1/4)/b+1/4*(c*x)^(7/2)*(b*x^2+a)^(1/4)/c+3 
/32*a^2*c^(5/2)*arctan(b^(1/4)*(c*x)^(1/2)/c^(1/2)/(b*x^2+a)^(1/4))/b^(7/4 
)-3/32*a^2*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 = 0.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.74 \[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\frac {(c x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2} \left (a+4 b x^2\right )+3 a^2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )-3 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{32 b^{7/4} x^{5/2}} \] Input:

Integrate[(c*x)^(5/2)*(a + b*x^2)^(1/4),x]
 

Output:

((c*x)^(5/2)*(2*b^(3/4)*x^(3/2)*(a + b*x^2)^(1/4)*(a + 4*b*x^2) + 3*a^2*Ar 
cTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] - 3*a^2*ArcTanh[(b^(1/4)*Sqrt[x] 
)/(a + b*x^2)^(1/4)]))/(32*b^(7/4)*x^(5/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {248, 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.

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

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{8} a \int \frac {(c x)^{5/2}}{\left (b x^2+a\right )^{3/4}}dx+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c^2 \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{3/4}}dx}{4 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c \int \frac {c x}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c \int \frac {c^3 x}{c^2-b c^2 x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c^3 \int \frac {c x}{c^2-b c^2 x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c^3 \left (\frac {\int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\sqrt {b} x c+c}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}\right )}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c^3 \left (\frac {\int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}\right )}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{8} a \left (\frac {c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b}-\frac {3 a c^3 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {c}}\right )}{2 b}\right )+\frac {(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}\)

Input:

Int[(c*x)^(5/2)*(a + b*x^2)^(1/4),x]
 

Output:

((c*x)^(7/2)*(a + b*x^2)^(1/4))/(4*c) + (a*((c*(c*x)^(3/2)*(a + b*x^2)^(1/ 
4))/(2*b) - (3*a*c^3*(-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])))/(2*b)))/8
 

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 248

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

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]

\[\int \left (c x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}d x\]

Input:

int((c*x)^(5/2)*(b*x^2+a)^(1/4),x)
 

Output:

int((c*x)^(5/2)*(b*x^2+a)^(1/4),x)
 

Fricas [F(-1)]

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

integrate((c*x)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.43 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.31 \[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt [4]{a} c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate((c*x)**(5/2)*(b*x**2+a)**(1/4),x)
 

Output:

a**(1/4)*c**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), b*x**2*e 
xp_polar(I*pi)/a)/(2*gamma(11/4))
 

Maxima [F]

\[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((c*x)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)*(c*x)^(5/2), x)
 

Giac [F]

\[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((c*x)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)*(c*x)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\int {\left (c\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \] Input:

int((c*x)^(5/2)*(a + b*x^2)^(1/4),x)
 

Output:

int((c*x)^(5/2)*(a + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt {c}\, c^{2} \left (2 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} a x +8 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{3}-3 \left (\int \frac {\sqrt {x}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2}\right )}{32 b} \] Input:

int((c*x)^(5/2)*(b*x^2+a)^(1/4),x)
 

Output:

(sqrt(c)*c**2*(2*sqrt(x)*(a + b*x**2)**(1/4)*a*x + 8*sqrt(x)*(a + b*x**2)* 
*(1/4)*b*x**3 - 3*int((sqrt(x)*(a + b*x**2)**(1/4))/(a + b*x**2),x)*a**2)) 
/(32*b)