3.9.63 \(\int \frac {(a+b x^2)^2}{(e x)^{5/2} (c+d x^2)^{5/2}} \, dx\) [863]

Optimal. Leaf size=258 \[ -\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt {e x}}{6 c^3 d e^3 \sqrt {c+d x^2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}} \]

[Out]

-2/3*a^2/c/e/(e*x)^(3/2)/(d*x^2+c)^(3/2)-1/3*(3*a^2*d^2-2*a*b*c*d+b^2*c^2)*(e*x)^(1/2)/c^2/d/e^3/(d*x^2+c)^(3/
2)+1/6*(b^2*c^2+5*a*d*(-3*a*d+2*b*c))*(e*x)^(1/2)/c^3/d/e^3/(d*x^2+c)^(1/2)+1/12*(b^2*c^2+5*a*d*(-3*a*d+2*b*c)
)*(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^(13/4)/d^(5/4)/e^(5/2)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {473, 468, 296, 335, 226} \begin {gather*} -\frac {\sqrt {e x} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (5 a d (2 b c-3 a d)+b^2 c^2\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}}+\frac {\sqrt {e x} \left (5 a d (2 b c-3 a d)+b^2 c^2\right )}{6 c^3 d e^3 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*(c + d*x^2)^(3/2)) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*Sqrt[e*x])/(3*c^2*d*e^3*(c
+ d*x^2)^(3/2)) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*d))*Sqrt[e*x])/(6*c^3*d*e^3*Sqrt[c + d*x^2]) + ((b^2*c^2 + 5
*a*d*(2*b*c - 3*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])/(12*c^(13/4)*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 296

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

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*} \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{5/2}} \, dx &=-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac {2 \int \frac {\frac {3}{2} a (2 b c-3 a d)+\frac {3}{2} b^2 c x^2}{\sqrt {e x} \left (c+d x^2\right )^{5/2}} \, dx}{3 c e^2}\\ &=-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \int \frac {1}{\sqrt {e x} \left (c+d x^2\right )^{3/2}} \, dx}{6 c^2 d e^2}\\ &=-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt {e x}}{6 c^3 d e^3 \sqrt {c+d x^2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{12 c^3 d e^2}\\ &=-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt {e x}}{6 c^3 d e^3 \sqrt {c+d x^2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{6 c^3 d e^3}\\ &=-\frac {2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt {e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt {e x}}{6 c^3 d e^3 \sqrt {c+d x^2}}+\frac {\left (b^2 c^2+5 a d (2 b c-3 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 )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.22, size = 211, normalized size = 0.82 \begin {gather*} \frac {x^{5/2} \left (\frac {b^2 c^2 x^2 \left (-c+d x^2\right )+2 a b c d x^2 \left (7 c+5 d x^2\right )-a^2 d \left (4 c^2+21 c d x^2+15 d^2 x^4\right )}{c^3 d x^{3/2} \left (c+d x^2\right )}+\frac {i \left (b^2 c^2+10 a b c d-15 a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{c^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} d}\right )}{6 (e x)^{5/2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*((b^2*c^2*x^2*(-c + d*x^2) + 2*a*b*c*d*x^2*(7*c + 5*d*x^2) - a^2*d*(4*c^2 + 21*c*d*x^2 + 15*d^2*x^4))
/(c^3*d*x^(3/2)*(c + d*x^2)) + (I*(b^2*c^2 + 10*a*b*c*d - 15*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSin
h[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(c^3*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(6*(e*x)^(5/2)*Sqrt[c + d*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(685\) vs. \(2(259)=518\).
time = 0.17, size = 686, normalized size = 2.66

method	result	size
elliptic	\(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d e \,x^{3}+c e x}}{3 c^{2} e^{3} d^{3} \left (x^{2}+\frac {c}{d}\right )^{2}}-\frac {x \left (11 a^{2} d^{2}-10 a b c d -b^{2} c^{2}\right )}{6 d \,e^{2} c^{3} \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{3 c^{3} e^{3} x^{2}}+\frac {\left (-\frac {11 a^{2} d^{2}-10 a b c d -b^{2} c^{2}}{12 d \,c^{3} e^{2}}-\frac {d \,a^{2}}{3 c^{3} e^{2}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(318\)
risch	\(-\frac {2 a^{2} \sqrt {d \,x^{2}+c}}{3 c^{3} x \,e^{2} \sqrt {e x}}-\frac {\left (\frac {a^{2} \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {d e \,x^{3}+c e x}}+\frac {3 c \left (a^{2} d^{2}-b^{2} c^{2}\right ) \left (\frac {x}{c \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {\sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{2 c d \sqrt {d e \,x^{3}+c e x}}\right )}{d}+\frac {3 c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {\sqrt {d e \,x^{3}+c e x}}{3 c e \,d^{2} \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {5 x}{6 c^{2} \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {5 \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{12 c^{2} d \sqrt {d e \,x^{3}+c e x}}\right )}{d}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{3 c^{3} e^{2} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(529\)
default	\(-\frac {15 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a^{2} d^{3} x^{3}-10 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a b c \,d^{2} x^{3}-\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, b^{2} c^{2} d \,x^{3}+15 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a^{2} c \,d^{2} x -10 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a b \,c^{2} d x -\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, b^{2} c^{3} x +30 a^{2} d^{4} x^{4}-20 a b c \,d^{3} x^{4}-2 b^{2} c^{2} d^{2} x^{4}+42 a^{2} c \,d^{3} x^{2}-28 a b \,c^{2} d^{2} x^{2}+2 b^{2} c^{3} d \,x^{2}+8 a^{2} c^{2} d^{2}}{12 x \,e^{2} \sqrt {e x}\, c^{3} d^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(686\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(15*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^
(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*a^2*d^3*x^3-10*((d*
x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)
*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*a*b*c*d^2*x^3-((d*x+(-c*d)^(1/2))
/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*
x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*b^2*c^2*d*x^3+15*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2
))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/
2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*a^2*c*d^2*x-10*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/
2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/
2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*a*b*c^2*d*x-((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^
(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(
1/2))*(-c*d)^(1/2)*b^2*c^3*x+30*a^2*d^4*x^4-20*a*b*c*d^3*x^4-2*b^2*c^2*d^2*x^4+42*a^2*c*d^3*x^2-28*a*b*c^2*d^2
*x^2+2*b^2*c^3*d*x^2+8*a^2*c^2*d^2)/x/e^2/(e*x)^(1/2)/c^3/d^2/(d*x^2+c)^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.28, size = 238, normalized size = 0.92 \begin {gather*} \frac {{\left ({\left ({\left (b^{2} c^{2} d^{2} + 10 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (b^{2} c^{3} d + 10 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{4} + {\left (b^{2} c^{4} + 10 \, a b c^{3} d - 15 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right ) - {\left (4 \, a^{2} c^{2} d^{2} - {\left (b^{2} c^{2} d^{2} + 10 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{4} + {\left (b^{2} c^{3} d - 14 \, a b c^{2} d^{2} + 21 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{6 \, {\left (c^{3} d^{4} x^{6} + 2 \, c^{4} d^{3} x^{4} + c^{5} d^{2} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(((b^2*c^2*d^2 + 10*a*b*c*d^3 - 15*a^2*d^4)*x^6 + 2*(b^2*c^3*d + 10*a*b*c^2*d^2 - 15*a^2*c*d^3)*x^4 + (b^2
*c^4 + 10*a*b*c^3*d - 15*a^2*c^2*d^2)*x^2)*sqrt(d)*weierstrassPInverse(-4*c/d, 0, x) - (4*a^2*c^2*d^2 - (b^2*c
^2*d^2 + 10*a*b*c*d^3 - 15*a^2*d^4)*x^4 + (b^2*c^3*d - 14*a*b*c^2*d^2 + 21*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c)*sqr
t(x))*e^(-5/2)/(c^3*d^4*x^6 + 2*c^4*d^3*x^4 + c^5*d^2*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{2}}{\left (e x\right )^{\frac {5}{2}} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{5/2}\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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