\(\int \frac {(a+b x^2)^{3/2}}{(c x)^{5/4}} \, dx\) [676]

Optimal result

Integrand size = 19, antiderivative size = 1045 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/4}} \, dx =\text {Too large to display} \] Output:

20/11*a*(b*x^2+a)^(1/2)/c/(c*x)^(1/4)+4/11*b*(c*x)^(7/4)*(b*x^2+a)^(1/2)/c 
^3+32/11*(2+2^(1/2))^(1/2)*a^(7/4)*b^(1/2)*(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))/c^(3/2)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2 
)+b^(1/4)*(c*x)^(1/2))+32/11*(2+2^(1/2))^(1/2)*a^(7/4)*b^(1/2)*(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))/c^(3/2)/(b*x^2+a)^( 
1/2)/(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))-32/11*a^(7/4)*b^(1/2)*(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) 
/c^(3/2)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-32/11*a^(7/ 
4)*b^(1/2)*(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)*Ell 
ipticF(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...
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.05 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/4}} \, dx=-\frac {4 a x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{8},\frac {7}{8},-\frac {b x^2}{a}\right )}{(c x)^{5/4} \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

(-4*a*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/2, -1/8, 7/8, -((b*x^2)/a)])/ 
((c*x)^(5/4)*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 = 0.25 (sec) , antiderivative size = 115, 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 = {247, 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.

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 248

\(\displaystyle \frac {12 b \left (\frac {4}{11} a \int \frac {(c x)^{3/4}}{\sqrt {b x^2+a}}dx+\frac {4 (c x)^{7/4} \sqrt {a+b x^2}}{11 c}\right )}{c^2}-\frac {4 \left (a+b x^2\right )^{3/2}}{c \sqrt [4]{c x}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {12 b \left (\frac {16 a \int \frac {(c x)^{3/2}}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{11 c}+\frac {4 (c x)^{7/4} \sqrt {a+b x^2}}{11 c}\right )}{c^2}-\frac {4 \left (a+b x^2\right )^{3/2}}{c \sqrt [4]{c x}}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {12 b \left (\frac {16 a \sqrt {\frac {b x^2}{a}+1} \int \frac {(c x)^{3/2}}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{11 c \sqrt {a+b x^2}}+\frac {4 (c x)^{7/4} \sqrt {a+b x^2}}{11 c}\right )}{c^2}-\frac {4 \left (a+b x^2\right )^{3/2}}{c \sqrt [4]{c x}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {12 b \left (\frac {16 a (c x)^{7/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},-\frac {b x^2}{a}\right )}{77 c \sqrt {a+b x^2}}+\frac {4 (c x)^{7/4} \sqrt {a+b x^2}}{11 c}\right )}{c^2}-\frac {4 \left (a+b x^2\right )^{3/2}}{c \sqrt [4]{c x}}\)

Input:

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

Output:

(-4*(a + b*x^2)^(3/2))/(c*(c*x)^(1/4)) + (12*b*((4*(c*x)^(7/4)*Sqrt[a + b* 
x^2])/(11*c) + (16*a*(c*x)^(7/4)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2 
, 7/8, 15/8, -((b*x^2)/a)])/(77*c*Sqrt[a + b*x^2])))/c^2
 

Defintions of rubi rules used

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.90 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.05 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/4}} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{8} \\ \frac {7}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{x} \Gamma \left (\frac {7}{8}\right )} \] Input:

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

Output:

a**(3/2)*gamma(-1/8)*hyper((-3/2, -1/8), (7/8,), b*x**2*exp_polar(I*pi)/a) 
/(2*c**(5/4)*x**(1/4)*gamma(7/8))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/4}} \, dx=\frac {\frac {20 \sqrt {b \,x^{2}+a}\, a}{11}+\frac {4 \sqrt {b \,x^{2}+a}\, b \,x^{2}}{11}+\frac {16 x^{\frac {1}{4}} \left (\int \frac {\sqrt {b \,x^{2}+a}}{x^{\frac {5}{4}} a +x^{\frac {13}{4}} b}d x \right ) a^{2}}{11}}{x^{\frac {1}{4}} c^{\frac {5}{4}}} \] Input:

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

Output:

(4*(5*sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2 + 4*x**(1/4)*int(sqrt(a 
 + b*x**2)/(x**(1/4)*a*x + x**(1/4)*b*x**3),x)*a**2))/(11*x**(1/4)*c**(1/4 
)*c)