\(\int \frac {(a+b x^2)^{3/4} (c+d x^2)}{x^2} \, dx\) [351]

Optimal result

Integrand size = 22, antiderivative size = 141 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^2} \, dx=\frac {3 (5 b c+2 a d) x}{5 \sqrt [4]{a+b x^2}}+\frac {(5 b c+2 a d) x \left (a+b x^2\right )^{3/4}}{5 a}-\frac {c \left (a+b x^2\right )^{7/4}}{a x}-\frac {3 \sqrt {a} (5 b c+2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {b} \sqrt [4]{a+b x^2}} \] Output:

3/5*(2*a*d+5*b*c)*x/(b*x^2+a)^(1/4)+1/5*(2*a*d+5*b*c)*x*(b*x^2+a)^(3/4)/a- 
c*(b*x^2+a)^(7/4)/a/x-3/5*a^(1/2)*(2*a*d+5*b*c)*(1+b*x^2/a)^(1/4)*Elliptic 
E(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(1/2)/(b*x^2+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 7.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^2} \, dx=\frac {\left (a+b x^2\right )^{3/4} \left (-c \left (a+b x^2\right )+\frac {(5 b c+2 a d) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{2 \left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{a x} \] Input:

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

Output:

((a + b*x^2)^(3/4)*(-(c*(a + b*x^2)) + ((5*b*c + 2*a*d)*x^2*Hypergeometric 
2F1[-3/4, 1/2, 3/2, -((b*x^2)/a)])/(2*(1 + (b*x^2)/a)^(3/4))))/(a*x)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {359, 211, 227, 225, 212}

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.

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {(2 a d+5 b c) \int \left (b x^2+a\right )^{3/4}dx}{2 a}-\frac {c \left (a+b x^2\right )^{7/4}}{a x}\)

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 227

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

\(\Big \downarrow \) 225

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

\(\Big \downarrow \) 212

\(\displaystyle \frac {(2 a d+5 b c) \left (\frac {3 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 \sqrt [4]{a+b x^2}}+\frac {2}{5} x \left (a+b x^2\right )^{3/4}\right )}{2 a}-\frac {c \left (a+b x^2\right )^{7/4}}{a x}\)

Input:

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

Output:

-((c*(a + b*x^2)^(7/4))/(a*x)) + ((5*b*c + 2*a*d)*((2*x*(a + b*x^2)^(3/4)) 
/5 + (3*a*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]* 
EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*(a + b*x^2)^(1/4 
))))/(2*a)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

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

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^2} \, dx=- \frac {a^{\frac {3}{4}} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + a^{\frac {3}{4}} d x {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \] Input:

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

Output:

-a**(3/4)*c*hyper((-3/4, -1/2), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + a**( 
3/4)*d*x*hyper((-3/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.57 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^2} \, dx=\frac {2\,c\,{\left (b\,x^2+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a}{b\,x^2}\right )}{x\,{\left (\frac {a}{b\,x^2}+1\right )}^{3/4}}+\frac {d\,x\,{\left (b\,x^2+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/4}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^2} \, dx=\frac {6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a d +10 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b c +2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b d \,x^{2}+6 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{4}+a \,x^{2}}d x \right ) a^{2} d x +15 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{4}+a \,x^{2}}d x \right ) a b c x}{5 b x} \] Input:

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

Output:

(6*(a + b*x**2)**(3/4)*a*d + 10*(a + b*x**2)**(3/4)*b*c + 2*(a + b*x**2)** 
(3/4)*b*d*x**2 + 6*int((a + b*x**2)**(3/4)/(a*x**2 + b*x**4),x)*a**2*d*x + 
 15*int((a + b*x**2)**(3/4)/(a*x**2 + b*x**4),x)*a*b*c*x)/(5*b*x)