\(\int \frac {(a+b x^2)^2}{(e x)^{5/2} (c+d x^2)^{3/2}} \, dx\) [855]

Optimal result

Integrand size = 28, antiderivative size = 207 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=-\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}-\frac {\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \sqrt {c+d x^2}}+\frac {\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \]

[Out]

-2/3*a^2/c/e/(e*x)^(3/2)/(d*x^2+c)^(1/2)-1/3*(5*a^2*d^2-6*a*b*c*d+3*b^2*c^2)*(e*x)^(1/2)/c^2/d/e^3/(d*x^2+c)^(
1/2)+1/6*(3*b^2*c^2+a*d*(-5*a*d+6*b*c))*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arc
tan(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^(9/4)/d^(5/4)/e^(5/2)/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {473, 468, 335, 226} \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=-\frac {\sqrt {e x} \left (5 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{3 c^2 d e^3 \sqrt {c+d x^2}}-\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (6 b c-5 a d)+3 b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \]

[In]

Int[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*Sqrt[c + d*x^2]) - ((3*b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*Sqrt[e*x])/(3*c^2*d*e^3*Sq
rt[c + d*x^2]) + ((3*b^2*c^2 + a*d*(6*b*c - 5*a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*
x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(6*c^(9/4)*d^(5/4)*e^(5/2)*Sqrt[c + d*x
^2])

Rule 226

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]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}+\frac {2 \int \frac {\frac {1}{2} a (6 b c-5 a d)+\frac {3}{2} b^2 c x^2}{\sqrt {e x} \left (c+d x^2\right )^{3/2}} \, dx}{3 c e^2} \\ & = -\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}-\frac {\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \sqrt {c+d x^2}}+\frac {\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{6 c^2 d e^2} \\ & = -\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}-\frac {\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \sqrt {c+d x^2}}+\frac {\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 c^2 d e^3} \\ & = -\frac {2 a^2}{3 c e (e x)^{3/2} \sqrt {c+d x^2}}-\frac {\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \sqrt {c+d x^2}}+\frac {\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.18 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\frac {x \left (-\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} \left (3 b^2 c^2 x^2-6 a b c d x^2+a^2 d \left (2 c+5 d x^2\right )\right )-i \left (-3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x^{5/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right ),-1\right )\right )}{3 c^2 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} d (e x)^{5/2} \sqrt {c+d x^2}} \]

[In]

Integrate[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(x*(-(Sqrt[(I*Sqrt[c])/Sqrt[d]]*(3*b^2*c^2*x^2 - 6*a*b*c*d*x^2 + a^2*d*(2*c + 5*d*x^2))) - I*(-3*b^2*c^2 - 6*a
*b*c*d + 5*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1]))/
(3*c^2*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d*(e*x)^(5/2)*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 3.68 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.29

method	result	size
elliptic	\(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {x \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{d \,e^{2} c^{2} \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{3 c^{2} e^{3} x^{2}}+\frac {\left (\frac {b^{2}}{e^{2} d}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 d \,c^{2} e^{2}}-\frac {d \,a^{2}}{3 c^{2} e^{2}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(268\)
risch	\(-\frac {2 a^{2} \sqrt {d \,x^{2}+c}}{3 c^{2} x \,e^{2} \sqrt {e x}}-\frac {\left (\frac {\left (a^{2} d^{2}-3 b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d^{2} \sqrt {d e \,x^{3}+c e x}}+\frac {3 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) c \left (\frac {x}{c \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {\sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{2 c d \sqrt {d e \,x^{3}+c e x}}\right )}{d}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{3 c^{2} e^{2} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(351\)
default	\(-\frac {5 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a^{2} d^{2} x -6 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a b c d x -3 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, b^{2} c^{2} x +10 a^{2} d^{3} x^{2}-12 a b c \,d^{2} x^{2}+6 b^{2} c^{2} d \,x^{2}+4 c \,a^{2} d^{2}}{6 x \sqrt {d \,x^{2}+c}\, c^{2} e^{2} \sqrt {e x}\, d^{2}}\)	\(353\)

[In]

int((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(e*x*(d*x^2+c))^(1/2)/(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-1/d/e^2*x/c^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/((x^2+c/d)*d*e*x
)^(1/2)-2/3/c^2/e^3*a^2*(d*e*x^3+c*e*x)^(1/2)/x^2+(b^2/e^2/d-1/2/d/c^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/e^2-1/3*d/c
^2/e^2*a^2)*(-c*d)^(1/2)/d*((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)))

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} - 5 \, a^{2} d^{3}\right )} x^{4} + {\left (3 \, b^{2} c^{3} + 6 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d e} {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right ) - {\left (2 \, a^{2} c d^{2} + {\left (3 \, b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{3 \, {\left (c^{2} d^{3} e^{3} x^{4} + c^{3} d^{2} e^{3} x^{2}\right )}} \]

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

1/3*(((3*b^2*c^2*d + 6*a*b*c*d^2 - 5*a^2*d^3)*x^4 + (3*b^2*c^3 + 6*a*b*c^2*d - 5*a^2*c*d^2)*x^2)*sqrt(d*e)*wei
erstrassPInverse(-4*c/d, 0, x) - (2*a^2*c*d^2 + (3*b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*x^2)*sqrt(d*x^2 + c)*s
qrt(e*x))/(c^2*d^3*e^3*x^4 + c^3*d^2*e^3*x^2)

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{\left (e x\right )^{\frac {5}{2}} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/((e*x)**(5/2)*(c + d*x**2)**(3/2)), x)

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(3/2)*(e*x)^(5/2)), x)

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(3/2)*(e*x)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{5/2}\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

[In]

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)), x)