\(\int \frac {(c x)^{5/4}}{\sqrt {a+b x^2}} \, dx\) [677]

Optimal result

Integrand size = 19, antiderivative size = 507 \[ \int \frac {(c x)^{5/4}}{\sqrt {a+b x^2}} \, dx=\frac {4 c \sqrt [4]{c x} \sqrt {a+b x^2}}{5 b}-\frac {2 a c (c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {\frac {\left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}-\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{5 \sqrt {2+\sqrt {2}} b^{3/4} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )}+\frac {2 a c (c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {-\frac {\left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}+\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{5 \sqrt {2+\sqrt {2}} b^{3/4} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )} \] Output:

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

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.14 \[ \int \frac {(c x)^{5/4}}{\sqrt {a+b x^2}} \, dx=\frac {4 c \sqrt [4]{c x} \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},-\frac {b x^2}{a}\right )\right )}{5 b \sqrt {a+b x^2}} \] Input:

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

Output:

(4*c*(c*x)^(1/4)*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/8, 
 1/2, 9/8, -((b*x^2)/a)]))/(5*b*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {262, 266, 767, 27, 2422}

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.

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {4 c \sqrt [4]{c x} \sqrt {a+b x^2}}{5 b}-\frac {4 a c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{5 b}\)

\(\Big \downarrow \) 767

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2422

\(\displaystyle \frac {4 c \sqrt [4]{c x} \sqrt {a+b x^2}}{5 b}-\frac {4 a c \left (\frac {\sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a c^2+b c^2 x^2}{\sqrt {a} \sqrt {b} c^2 x}} \sqrt {\frac {\left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {b} x c+\sqrt {2} \sqrt {a} c-2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {c x} \sqrt {c}}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )}-\frac {\sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a c^2+b c^2 x^2}{\sqrt {a} \sqrt {b} c^2 x}} \sqrt {-\frac {\left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {b} x c+\sqrt {2} \sqrt {a} c+2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {c x} \sqrt {c}}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )}\right )}{5 b}\)

Input:

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

Output:

(4*c*(c*x)^(1/4)*Sqrt[a + b*x^2])/(5*b) - (4*a*c*((b^(1/4)*(c*x)^(3/4)*Sqr 
t[-((a*c^2 + b*c^2*x^2)/(Sqrt[a]*Sqrt[b]*c^2*x))]*Sqrt[(a^(1/4)*Sqrt[c] + 
b^(1/4)*Sqrt[c*x])^2/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x])]*EllipticF[ArcSin 
[Sqrt[-((Sqrt[2]*Sqrt[a]*c + Sqrt[2]*Sqrt[b]*c*x - 2*a^(1/4)*b^(1/4)*Sqrt[ 
c]*Sqrt[c*x])/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x]))]/2], -2*(1 - Sqrt[2])]) 
/(2*Sqrt[2 + Sqrt[2]]*Sqrt[a + b*x^2]*(a^(1/4)*Sqrt[c] + b^(1/4)*Sqrt[c*x] 
)) - (b^(1/4)*(c*x)^(3/4)*Sqrt[-((a*c^2 + b*c^2*x^2)/(Sqrt[a]*Sqrt[b]*c^2* 
x))]*Sqrt[-((a^(1/4)*Sqrt[c] - b^(1/4)*Sqrt[c*x])^2/(a^(1/4)*b^(1/4)*Sqrt[ 
c]*Sqrt[c*x]))]*EllipticF[ArcSin[Sqrt[(Sqrt[2]*Sqrt[a]*c + Sqrt[2]*Sqrt[b] 
*c*x + 2*a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x])/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[ 
c*x])]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*Sqrt[a + b*x^2]*(a^(1/4 
)*Sqrt[c] - b^(1/4)*Sqrt[c*x]))))/(5*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 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 767

Int[1/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[1/2   Int[(1 - Rt[b/a, 4 
]*x^2)/Sqrt[a + b*x^8], x], x] + Simp[1/2   Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a 
 + b*x^8], x], x] /; FreeQ[{a, b}, x]
 

rule 2422

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[(-c) 
*d*x^3*Sqrt[-(c - d*x^2)^2/(c*d*x^2)]*(Sqrt[(-d^2)*((a + b*x^8)/(b*c^2*x^4) 
)]/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]))*EllipticF[ArcSin[(1/2)* 
Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4)/(c*d*x^2)]], -2*(1 - Sqrt[ 
2])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^4 - a*d^4, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.63 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.09 \[ \int \frac {(c x)^{5/4}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {5}{4}} x^{\frac {9}{4}} \Gamma \left (\frac {9}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{8} \\ \frac {17}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {17}{8}\right )} \] Input:

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

Output:

c**(5/4)*x**(9/4)*gamma(9/8)*hyper((1/2, 9/8), (17/8,), b*x**2*exp_polar(I 
*pi)/a)/(2*sqrt(a)*gamma(17/8))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c x)^{5/4}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {5}{4}} \left (4 x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}-\left (\int \frac {\sqrt {b \,x^{2}+a}}{x^{\frac {3}{4}} a +x^{\frac {11}{4}} b}d x \right ) a \right )}{5 b} \] Input:

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

Output:

(c**(1/4)*c*(4*x**(1/4)*sqrt(a + b*x**2) - int(sqrt(a + b*x**2)/(x**(3/4)* 
a + x**(3/4)*b*x**2),x)*a))/(5*b)