\(\int \frac {(a+b x^2)^{3/4} (c+d x^2)}{x^4} \, dx\) [352]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^4} \, dx=\frac {b (b c+6 a d) x}{2 a \sqrt [4]{a+b x^2}}-\frac {(b c+6 a d) \left (a+b x^2\right )^{3/4}}{6 a x}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}-\frac {\sqrt {b} (b c+6 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a+b x^2}} \] Output:

1/2*b*(6*a*d+b*c)*x/a/(b*x^2+a)^(1/4)-1/6*(6*a*d+b*c)*(b*x^2+a)^(3/4)/a/x- 
1/3*c*(b*x^2+a)^(7/4)/a/x^3-1/2*b^(1/2)*(6*a*d+b*c)*(1+b*x^2/a)^(1/4)*Elli 
pticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(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 = 7.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^4} \, dx=\frac {\left (a+b x^2\right )^{3/4} \left (-c \left (a+b x^2\right )-\frac {(b c+6 a d) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{2},-\frac {b x^2}{a}\right )}{2 \left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{3 a x^3} \] Input:

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

Output:

((a + b*x^2)^(3/4)*(-(c*(a + b*x^2)) - ((b*c + 6*a*d)*x^2*Hypergeometric2F 
1[-3/4, -1/2, 1/2, -((b*x^2)/a)])/(2*(1 + (b*x^2)/a)^(3/4))))/(3*a*x^3)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {359, 247, 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.

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {(6 a d+b c) \int \frac {\left (b x^2+a\right )^{3/4}}{x^2}dx}{6 a}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {(6 a d+b c) \left (\frac {3}{2} b \int \frac {1}{\sqrt [4]{b x^2+a}}dx-\frac {\left (a+b x^2\right )^{3/4}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {(6 a d+b c) \left (\frac {3 b \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {(6 a d+b c) \left (\frac {3 b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(6 a d+b c) \left (\frac {3 b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{7/4}}{3 a x^3}\)

Input:

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

Output:

-1/3*(c*(a + b*x^2)^(7/4))/(a*x^3) + ((b*c + 6*a*d)*(-((a + b*x^2)^(3/4)/x 
) + (3*b*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*E 
llipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(2*(a + b*x^2)^(1/4) 
)))/(6*a)
 

Defintions of rubi rules used

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 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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.45 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^4} \, dx=- \frac {a^{\frac {3}{4}} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} - \frac {a^{\frac {3}{4}} d {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{x^4} \, dx=\frac {-6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a d -2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b c +6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b d \,x^{2}-18 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{6}+a \,x^{4}}d x \right ) a^{2} d \,x^{3}-3 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{6}+a \,x^{4}}d x \right ) a b c \,x^{3}}{3 b \,x^{3}} \] Input:

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

Output:

( - 6*(a + b*x**2)**(3/4)*a*d - 2*(a + b*x**2)**(3/4)*b*c + 6*(a + b*x**2) 
**(3/4)*b*d*x**2 - 18*int((a + b*x**2)**(3/4)/(a*x**4 + b*x**6),x)*a**2*d* 
x**3 - 3*int((a + b*x**2)**(3/4)/(a*x**4 + b*x**6),x)*a*b*c*x**3)/(3*b*x** 
3)