\(\int \sqrt [4]{c x} (a+b x^2)^{3/2} \, dx\) [673]

Optimal result

Integrand size = 19, antiderivative size = 1024 \[ \int \sqrt [4]{c x} \left (a+b x^2\right )^{3/2} \, dx =\text {Too large to display} \] Output:

16/51*a*(c*x)^(5/4)*(b*x^2+a)^(1/2)/c+4/17*(c*x)^(5/4)*(b*x^2+a)^(3/2)/c-3 
2/51*(2+2^(1/2))^(1/2)*a^(5/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^ 
(1/2)*((a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c* 
x)^(1/2))^(1/2)*EllipticE(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x 
/a^(1/2)-2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2) 
,(-2+2*2^(1/2))^(1/2))/b^(1/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c 
*x)^(1/2))+32/51*(2+2^(1/2))^(1/2)*a^(5/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2) 
/b^(1/2)/x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4 
)/c^(1/2)/(c*x)^(1/2))^(1/2)*EllipticE(1/2*(a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/ 
2)*b^(1/2)*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^ 
(1/2))^(1/2),(-2+2*2^(1/2))^(1/2))/b^(1/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2 
)-b^(1/4)*(c*x)^(1/2))+32/51*a^(5/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/ 
2)/x)^(1/2)*((a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/ 
2)/(c*x)^(1/2))^(1/2)*EllipticF(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^( 
1/2)*x/a^(1/2)-2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2)) 
^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/b^(1/4)/(b*x^2+a)^(1/2)/(a^ 
(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-32/51*a^(5/2)*(c*x)^(3/4)*(-(b*x^2+a)/a 
^(1/2)/b^(1/2)/x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/ 
b^(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2)*EllipticF(1/2*(a^(1/4)*c^(1/2)*(2^(1/2) 
+2^(1/2)*b^(1/2)*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/...
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.06 \[ \int \sqrt [4]{c x} \left (a+b x^2\right )^{3/2} \, dx=\frac {4 a x \sqrt [4]{c x} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{5 \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

(4*a*x*(c*x)^(1/4)*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/2, 5/8, 13/8, -((b 
*x^2)/a)])/(5*Sqrt[1 + (b*x^2)/a])
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {248, 248, 266, 889, 888}

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.

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

\(\Big \downarrow \) 248

\(\displaystyle \frac {12}{17} a \int \sqrt [4]{c x} \sqrt {b x^2+a}dx+\frac {4 (c x)^{5/4} \left (a+b x^2\right )^{3/2}}{17 c}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {12}{17} a \left (\frac {4}{9} a \int \frac {\sqrt [4]{c x}}{\sqrt {b x^2+a}}dx+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\right )+\frac {4 (c x)^{5/4} \left (a+b x^2\right )^{3/2}}{17 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {12}{17} a \left (\frac {16 a \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{9 c}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\right )+\frac {4 (c x)^{5/4} \left (a+b x^2\right )^{3/2}}{17 c}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {12}{17} a \left (\frac {16 a \sqrt {\frac {b x^2}{a}+1} \int \frac {c x}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{9 c \sqrt {a+b x^2}}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\right )+\frac {4 (c x)^{5/4} \left (a+b x^2\right )^{3/2}}{17 c}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {12}{17} a \left (\frac {16 a (c x)^{5/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{45 c \sqrt {a+b x^2}}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\right )+\frac {4 (c x)^{5/4} \left (a+b x^2\right )^{3/2}}{17 c}\)

Input:

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

Output:

(4*(c*x)^(5/4)*(a + b*x^2)^(3/2))/(17*c) + (12*a*((4*(c*x)^(5/4)*Sqrt[a + 
b*x^2])/(9*c) + (16*a*(c*x)^(5/4)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/ 
2, 5/8, 13/8, -((b*x^2)/a)])/(45*c*Sqrt[a + b*x^2])))/17
 

Defintions of rubi rules used

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 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 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 889

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I 
ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(c*x) 
^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0 
] &&  !(ILtQ[p, 0] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.04 \[ \int \sqrt [4]{c x} \left (a+b x^2\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} \sqrt [4]{c} x^{\frac {5}{4}} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{8} \\ \frac {13}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{8}\right )} \] Input:

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

Output:

a**(3/2)*c**(1/4)*x**(5/4)*gamma(5/8)*hyper((-3/2, 5/8), (13/8,), b*x**2*e 
xp_polar(I*pi)/a)/(2*gamma(13/8))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{c x} \left (a+b x^2\right )^{3/2} \, dx=\frac {4 c^{\frac {1}{4}} \left (7 x^{\frac {5}{4}} \sqrt {b \,x^{2}+a}\, a +3 x^{\frac {13}{4}} \sqrt {b \,x^{2}+a}\, b +4 \left (\int \frac {x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}}{b \,x^{2}+a}d x \right ) a^{2}\right )}{51} \] Input:

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

Output:

(4*c**(1/4)*(7*x**(1/4)*sqrt(a + b*x**2)*a*x + 3*x**(1/4)*sqrt(a + b*x**2) 
*b*x**3 + 4*int((x**(1/4)*sqrt(a + b*x**2))/(a + b*x**2),x)*a**2))/51