\(\int \frac {(a+b x^2)^{3/4} (c+d x^2)}{(e x)^{11/2}} \, dx\) [436]

Optimal result

Integrand size = 26, antiderivative size = 182 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=\frac {2 b (2 b c-9 a d)}{15 a e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {2 (2 b c-9 a d) \left (a+b x^2\right )^{3/4}}{45 a e^3 (e x)^{5/2}}-\frac {2 c \left (a+b x^2\right )^{7/4}}{9 a e (e x)^{9/2}}-\frac {2 b^{3/2} (2 b c-9 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{3/2} e^6 \sqrt [4]{a+b x^2}} \] Output:

2/15*b*(-9*a*d+2*b*c)/a/e^5/(e*x)^(1/2)/(b*x^2+a)^(1/4)+2/45*(-9*a*d+2*b*c 
)*(b*x^2+a)^(3/4)/a/e^3/(e*x)^(5/2)-2/9*c*(b*x^2+a)^(7/4)/a/e/(e*x)^(9/2)- 
2/15*b^(3/2)*(-9*a*d+2*b*c)*(1+a/b/x^2)^(1/4)*(e*x)^(1/2)*EllipticE(sin(1/ 
2*arccot(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(3/2)/e^6/(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 = 10.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=\frac {2 x \left (a+b x^2\right )^{3/4} \left (-5 c \left (a+b x^2\right )+\frac {(2 b c-9 a d) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {3}{4},-\frac {1}{4},-\frac {b x^2}{a}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{45 a (e x)^{11/2}} \] Input:

Integrate[((a + b*x^2)^(3/4)*(c + d*x^2))/(e*x)^(11/2),x]
 

Output:

(2*x*(a + b*x^2)^(3/4)*(-5*c*(a + b*x^2) + ((2*b*c - 9*a*d)*x^2*Hypergeome 
tric2F1[-5/4, -3/4, -1/4, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(3/4)))/(45*a*(e* 
x)^(11/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {359, 247, 257, 249, 858, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx\)

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 257

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

\(\Big \downarrow \) 249

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 212

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

Input:

Int[((a + b*x^2)^(3/4)*(c + d*x^2))/(e*x)^(11/2),x]
 

Output:

(-2*c*(a + b*x^2)^(7/4))/(9*a*e*(e*x)^(9/2)) - ((2*b*c - 9*a*d)*((-2*(a + 
b*x^2)^(3/4))/(5*e*(e*x)^(5/2)) + (3*b*(-2/(e*Sqrt[e*x]*(a + b*x^2)^(1/4)) 
 + (2*Sqrt[b]*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sq 
rt[b]*x)]/2, 2])/(Sqrt[a]*e^2*(a + b*x^2)^(1/4))))/(5*e^2)))/(9*a*e^2)
 

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 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 249

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

rule 257

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Simp[- 
2/(c*Sqrt[c*x]*(a + b*x^2)^(1/4)), x] - Simp[b/c^2   Int[Sqrt[c*x]/(a + b*x 
^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/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]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )}{\left (e x \right )^{\frac {11}{2}}}d x\]

Input:

int((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x)
 

Output:

int((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x)
 

Fricas [F]

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

integrate((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(e^6*x^6), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 100.76 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=- \frac {b^{\frac {3}{4}} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{3 e^{\frac {11}{2}} x^{3}} - \frac {b^{\frac {3}{4}} d {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{e^{\frac {11}{2}} x} \] Input:

integrate((b*x**2+a)**(3/4)*(d*x**2+c)/(e*x)**(11/2),x)
 

Output:

-b**(3/4)*c*hyper((-3/4, 3/2), (5/2,), a*exp_polar(I*pi)/(b*x**2))/(3*e**( 
11/2)*x**3) - b**(3/4)*d*hyper((-3/4, 1/2), (3/2,), a*exp_polar(I*pi)/(b*x 
**2))/(e**(11/2)*x)
 

Maxima [F]

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

integrate((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(3/4)*(d*x^2 + c)/(e*x)^(11/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (d\,x^2+c\right )}{{\left (e\,x\right )}^{11/2}} \,d x \] Input:

int(((a + b*x^2)^(3/4)*(c + d*x^2))/(e*x)^(11/2),x)
 

Output:

int(((a + b*x^2)^(3/4)*(c + d*x^2))/(e*x)^(11/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{11/2}} \, dx=\frac {\sqrt {e}\, \left (6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a d -4 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b c -12 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b d \,x^{2}+27 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{8}+a \,x^{6}}d x \right ) a^{2} d \,x^{4}-6 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{8}+a \,x^{6}}d x \right ) a b c \,x^{4}\right )}{12 \sqrt {x}\, b \,e^{6} x^{4}} \] Input:

int((b*x^2+a)^(3/4)*(d*x^2+c)/(e*x)^(11/2),x)
 

Output:

(sqrt(e)*(6*(a + b*x**2)**(3/4)*a*d - 4*(a + b*x**2)**(3/4)*b*c - 12*(a + 
b*x**2)**(3/4)*b*d*x**2 + 27*sqrt(x)*int((sqrt(x)*(a + b*x**2)**(3/4))/(a* 
x**6 + b*x**8),x)*a**2*d*x**4 - 6*sqrt(x)*int((sqrt(x)*(a + b*x**2)**(3/4) 
)/(a*x**6 + b*x**8),x)*a*b*c*x**4))/(12*sqrt(x)*b*e**6*x**4)