3.9.44 \(\int \frac {(a+b x^2)^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx\) [844]

3.9.44.1 Optimal result
3.9.44.2 Mathematica [C] (verified)
3.9.44.3 Rubi [A] (verified)
3.9.44.4 Maple [A] (verified)
3.9.44.5 Fricas [C] (verification not implemented)
3.9.44.6 Sympy [C] (verification not implemented)
3.9.44.7 Maxima [F]
3.9.44.8 Giac [F]
3.9.44.9 Mupad [F(-1)]

3.9.44.1 Optimal result

Integrand size = 28, antiderivative size = 184 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx=-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3}-\frac {\left (b^2 c^2-6 a b c d+a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 c^{5/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \]

output
-2/3*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(3/2)+2/3*b^2*(e*x)^(1/2)*(d*x^2+c)^(1/ 
2)/d/e^3-1/3*(a^2*d^2-6*a*b*c*d+b^2*c^2)*(cos(2*arctan(d^(1/4)*(e*x)^(1/2) 
/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/ 
2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1 
/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^(5/4)/d 
^(5/4)/e^(5/2)/(d*x^2+c)^(1/2)
 
3.9.44.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx=\frac {x \left (2 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} \left (-a^2 d+b^2 c x^2\right ) \left (c+d x^2\right )-2 i \left (b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x^{5/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right ),-1\right )\right )}{3 c \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} d (e x)^{5/2} \sqrt {c+d x^2}} \]

input
Integrate[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x^2]),x]
 
output
(x*(2*Sqrt[(I*Sqrt[c])/Sqrt[d]]*(-(a^2*d) + b^2*c*x^2)*(c + d*x^2) - (2*I) 
*(b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^(5/2)*EllipticF[I*A 
rcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1]))/(3*c*Sqrt[(I*Sqrt[c])/Sqr 
t[d]]*d*(e*x)^(5/2)*Sqrt[c + d*x^2])
 
3.9.44.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {365, 27, 363, 266, 761}

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.

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

\(\Big \downarrow \) 365

\(\displaystyle \frac {2 \int \frac {3 b^2 c x^2+a (6 b c-a d)}{2 \sqrt {e x} \sqrt {d x^2+c}}dx}{3 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 b^2 c x^2+a (6 b c-a d)}{\sqrt {e x} \sqrt {d x^2+c}}dx}{3 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}\)

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {\frac {2 b^2 c \sqrt {e x} \sqrt {c+d x^2}}{d e}-\frac {\left (a^2 d^2-6 a b c d+b^2 c^2\right ) \left (\sqrt {c} e+\sqrt {d} e x\right ) \sqrt {\frac {c e^2+d e^2 x^2}{\left (\sqrt {c} e+\sqrt {d} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{\sqrt [4]{c} d^{5/4} e^{3/2} \sqrt {c+d x^2}}}{3 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}\)

input
Int[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x^2]),x]
 
output
(-2*a^2*Sqrt[c + d*x^2])/(3*c*e*(e*x)^(3/2)) + ((2*b^2*c*Sqrt[e*x]*Sqrt[c 
+ d*x^2])/(d*e) - ((b^2*c^2 - 6*a*b*c*d + a^2*d^2)*(Sqrt[c]*e + Sqrt[d]*e* 
x)*Sqrt[(c*e^2 + d*e^2*x^2)/(Sqrt[c]*e + Sqrt[d]*e*x)^2]*EllipticF[2*ArcTa 
n[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(c^(1/4)*d^(5/4)*e^(3/2)*S 
qrt[c + d*x^2]))/(3*c*e^2)
 

3.9.44.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 363
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 365
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

rule 761
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 
3.9.44.4 Maple [A] (verified)

Time = 3.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.13

method result size
risch \(-\frac {2 \sqrt {d \,x^{2}+c}\, \left (-b^{2} c \,x^{2}+a^{2} d \right )}{3 d c x \,e^{2} \sqrt {e x}}-\frac {\left (a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{3 c \,d^{2} \sqrt {d e \,x^{3}+c e x}\, e^{2} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\) \(208\)
elliptic \(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{3 e^{3} c \,x^{2}}+\frac {2 b^{2} \sqrt {d e \,x^{3}+c e x}}{3 e^{3} d}+\frac {\left (\frac {2 a b}{e^{2}}-\frac {d \,a^{2}}{3 c \,e^{2}}-\frac {b^{2} c}{3 e^{2} d}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\) \(221\)
default \(-\frac {\sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a^{2} d^{2} x -6 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a b c d x +\sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, b^{2} c^{2} x -2 b^{2} c \,d^{2} x^{4}+2 a^{2} d^{3} x^{2}-2 b^{2} c^{2} d \,x^{2}+2 c \,a^{2} d^{2}}{3 \sqrt {d \,x^{2}+c}\, x c \,e^{2} \sqrt {e x}\, d^{2}}\) \(352\)

input
int((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
 
output
-2/3*(d*x^2+c)^(1/2)*(-b^2*c*x^2+a^2*d)/d/c/x/e^2/(e*x)^(1/2)-1/3*(a^2*d^2 
-6*a*b*c*d+b^2*c^2)/c/d^2*(-c*d)^(1/2)*((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d) 
^(1/2)*(-2*(x-(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2)*(-x/(-c*d)^(1/2)*d)^(1 
/2)/(d*e*x^3+c*e*x)^(1/2)*EllipticF(((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1 
/2),1/2*2^(1/2))/e^2*(e*x*(d*x^2+c))^(1/2)/(e*x)^(1/2)/(d*x^2+c)^(1/2)
 
3.9.44.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.48 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx=-\frac {2 \, {\left ({\left (b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \sqrt {d e} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right ) - {\left (b^{2} c d x^{2} - a^{2} d^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}\right )}}{3 \, c d^{2} e^{3} x^{2}} \]

input
integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")
 
output
-2/3*((b^2*c^2 - 6*a*b*c*d + a^2*d^2)*sqrt(d*e)*x^2*weierstrassPInverse(-4 
*c/d, 0, x) - (b^2*c*d*x^2 - a^2*d^2)*sqrt(d*x^2 + c)*sqrt(e*x))/(c*d^2*e^ 
3*x^2)
 
3.9.44.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.71 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx=\frac {a^{2} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {a b \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{2} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} \]

input
integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(1/2),x)
 
output
a**2*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), d*x**2*exp_polar(I*pi)/c)/(2*s 
qrt(c)*e**(5/2)*x**(3/2)*gamma(1/4)) + a*b*sqrt(x)*gamma(1/4)*hyper((1/4, 
1/2), (5/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*e**(5/2)*gamma(5/4)) + b* 
*2*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c) 
/(2*sqrt(c)*e**(5/2)*gamma(9/4))
 
3.9.44.7 Maxima [F]

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

input
integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")
 
output
integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(5/2)), x)
 
3.9.44.8 Giac [F]

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

input
integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")
 
output
integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(5/2)), x)
 
3.9.44.9 Mupad [F(-1)]

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

input
int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(1/2)),x)
 
output
int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(1/2)), x)