\(\int (a+b x^2)^{3/4} (c+d x^2) \, dx\) [412]

Optimal result

Integrand size = 19, antiderivative size = 145 \[ \int \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 a (9 b c-2 a d) x}{15 b \sqrt [4]{a+b x^2}}+\frac {2 (9 b c-2 a d) x \left (a+b x^2\right )^{3/4}}{45 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}-\frac {2 a^{3/2} (9 b c-2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^2}} \] Output:

2/15*a*(-2*a*d+9*b*c)*x/b/(b*x^2+a)^(1/4)+2/45*(-2*a*d+9*b*c)*x*(b*x^2+a)^ 
(3/4)/b+2/9*d*x*(b*x^2+a)^(7/4)/b-2/15*a^(3/2)*(-2*a*d+9*b*c)*(1+b*x^2/a)^ 
(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(3/2)/(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 = 7.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 x \left (a+b x^2\right )^{3/4} \left (d \left (a+b x^2\right )+\frac {(9 b c-2 a d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{2 \left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{9 b} \] Input:

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

Output:

(2*x*(a + b*x^2)^(3/4)*(d*(a + b*x^2) + ((9*b*c - 2*a*d)*Hypergeometric2F1 
[-3/4, 1/2, 3/2, -((b*x^2)/a)])/(2*(1 + (b*x^2)/a)^(3/4))))/(9*b)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {299, 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.

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {(9 b c-2 a d) \int \left (b x^2+a\right )^{3/4}dx}{9 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(9 b c-2 a d) \left (\frac {3}{5} a \int \frac {1}{\sqrt [4]{b x^2+a}}dx+\frac {2}{5} x \left (a+b x^2\right )^{3/4}\right )}{9 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {(9 b c-2 a d) \left (\frac {3 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 \sqrt [4]{a+b x^2}}+\frac {2}{5} x \left (a+b x^2\right )^{3/4}\right )}{9 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {(9 b c-2 a d) \left (\frac {3 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 \sqrt [4]{a+b x^2}}+\frac {2}{5} x \left (a+b x^2\right )^{3/4}\right )}{9 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(9 b c-2 a d) \left (\frac {3 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 \sqrt [4]{a+b x^2}}+\frac {2}{5} x \left (a+b x^2\right )^{3/4}\right )}{9 b}+\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 b}\)

Input:

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

Output:

(2*d*x*(a + b*x^2)^(7/4))/(9*b) + ((9*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))))/(9*b)
 

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 299

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.41 \[ \int \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=a^{\frac {3}{4}} c x {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{\frac {3}{4}} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:

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

Output:

a**(3/4)*c*x*hyper((-3/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**(3/ 
4)*d*x**3*hyper((-3/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a d x +18 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b c x +10 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b d \,x^{3}-6 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a^{2} d +27 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a b c}{45 b} \] Input:

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

Output:

(6*(a + b*x**2)**(3/4)*a*d*x + 18*(a + b*x**2)**(3/4)*b*c*x + 10*(a + b*x* 
*2)**(3/4)*b*d*x**3 - 6*int((a + b*x**2)**(3/4)/(a + b*x**2),x)*a**2*d + 2 
7*int((a + b*x**2)**(3/4)/(a + b*x**2),x)*a*b*c)/(45*b)