3.213 \(\int \frac {1}{(a+b x^2)^{5/2} (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=323 \[ -\frac {b \sqrt {c} \sqrt {d} \sqrt {a+b x^2} (b c-9 a d) \operatorname {EllipticF}\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 a^2 \sqrt {c+d x^2} (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\sqrt {d} \sqrt {a+b x^2} \left (-3 a^2 d^2-7 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a^2 \sqrt {c} \sqrt {c+d x^2} (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {2 b x (b c-3 a d)}{3 a^2 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2}+\frac {b x}{3 a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d)} \]

[Out]

1/3*b*x/a/(-a*d+b*c)/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2)+2/3*b*(-3*a*d+b*c)*x/a^2/(-a*d+b*c)^2/(b*x^2+a)^(1/2)/(d*
x^2+c)^(1/2)+1/3*(-3*a^2*d^2-7*a*b*c*d+2*b^2*c^2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/
c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*d^(1/2)*(b*x^2+a)^(1/2)/a^2/(-a*d+b*c)^3/c^(1/2)/(c*(b*x^2+a)/a/(
d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/3*b*(-9*a*d+b*c)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)
/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*d^(1/2)*(b*x^2+a)^(1/2)/a^2/(-a*d+b*c)^3/(c*(b*x^2+a)/a/
(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {414, 527, 525, 418, 411} \[ \frac {\sqrt {d} \sqrt {a+b x^2} \left (-3 a^2 d^2-7 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a^2 \sqrt {c} \sqrt {c+d x^2} (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {2 b x (b c-3 a d)}{3 a^2 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2}-\frac {b \sqrt {c} \sqrt {d} \sqrt {a+b x^2} (b c-9 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a^2 \sqrt {c+d x^2} (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {b x}{3 a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(b*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2]) + (2*b*(b*c - 3*a*d)*x)/(3*a^2*(b*c - a*d)^2*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2]) + (Sqrt[d]*(2*b^2*c^2 - 7*a*b*c*d - 3*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqr
t[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*a^2*Sqrt[c]*(b*c - a*d)^3*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c
+ d*x^2]) - (b*Sqrt[c]*Sqrt[d]*(b*c - 9*a*d)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/
(a*d)])/(3*a^2*(b*c - a*d)^3*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 411

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^{3/2}} \, dx &=\frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}-\frac {\int \frac {-2 b c+3 a d-3 b d x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^{3/2}} \, dx}{3 a (b c-a d)}\\ &=\frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}+\frac {2 b (b c-3 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}+\frac {\int \frac {a d (b c+3 a d)+2 b d (b c-3 a d) x^2}{\sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \, dx}{3 a^2 (b c-a d)^2}\\ &=\frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}+\frac {2 b (b c-3 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}-\frac {(b d (b c-9 a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 a (b c-a d)^3}+\frac {\left (d \left (2 b^2 c^2-7 a b c d-3 a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 a^2 (b c-a d)^3}\\ &=\frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}+\frac {2 b (b c-3 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}+\frac {\sqrt {d} \left (2 b^2 c^2-7 a b c d-3 a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a^2 \sqrt {c} (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {b \sqrt {c} \sqrt {d} (b c-9 a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a^2 (b c-a d)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.07, size = 337, normalized size = 1.04 \[ \frac {2 i b c \left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (3 a^2 d^2-4 a b c d+b^2 c^2\right ) \operatorname {EllipticF}\left (i \sinh ^{-1}\left (x \sqrt {\frac {b}{a}}\right ),\frac {a d}{b c}\right )+i b c \left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (3 a^2 d^2+7 a b c d-2 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+x \sqrt {\frac {b}{a}} \left (3 a^4 d^3+6 a^3 b d^3 x^2+a^2 b^2 d \left (8 c^2+8 c d x^2+3 d^2 x^4\right )+a b^3 c \left (-3 c^2+4 c d x^2+7 d^2 x^4\right )-2 b^4 c^2 x^2 \left (c+d x^2\right )\right )}{3 a^2 c \sqrt {\frac {b}{a}} \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^2)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(Sqrt[b/a]*x*(3*a^4*d^3 + 6*a^3*b*d^3*x^2 - 2*b^4*c^2*x^2*(c + d*x^2) + a^2*b^2*d*(8*c^2 + 8*c*d*x^2 + 3*d^2*x
^4) + a*b^3*c*(-3*c^2 + 4*c*d*x^2 + 7*d^2*x^4)) + I*b*c*(-2*b^2*c^2 + 7*a*b*c*d + 3*a^2*d^2)*(a + b*x^2)*Sqrt[
1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (2*I)*b*c*(b^2*c^2 - 4*a*b
*c*d + 3*a^2*d^2)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/
(b*c)])/(3*a^2*Sqrt[b/a]*c*(-(b*c) + a*d)^3*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{b^{3} d^{2} x^{10} + {\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{8} + {\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{6} + a^{3} c^{2} + {\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{4} + {\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(b^3*d^2*x^10 + (2*b^3*c*d + 3*a*b^2*d^2)*x^8 + (b^3*c^2 + 6*a*b^2*c*
d + 3*a^2*b*d^2)*x^6 + a^3*c^2 + (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*x^4 + (3*a^2*b*c^2 + 2*a^3*c*d)*x^2), x
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(d*x^2 + c)^(3/2)), x)

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maple [B]  time = 0.07, size = 964, normalized size = 2.98 \[ -\frac {-3 \sqrt {-\frac {b}{a}}\, a^{2} b^{2} d^{3} x^{5}-7 \sqrt {-\frac {b}{a}}\, a \,b^{3} c \,d^{2} x^{5}+2 \sqrt {-\frac {b}{a}}\, b^{4} c^{2} d \,x^{5}-6 \sqrt {-\frac {b}{a}}\, a^{3} b \,d^{3} x^{3}-8 \sqrt {-\frac {b}{a}}\, a^{2} b^{2} c \,d^{2} x^{3}+3 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{2} b^{2} c \,d^{2} x^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+6 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{2} b^{2} c \,d^{2} x^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-4 \sqrt {-\frac {b}{a}}\, a \,b^{3} c^{2} d \,x^{3}+7 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a \,b^{3} c^{2} d \,x^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a \,b^{3} c^{2} d \,x^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {-\frac {b}{a}}\, b^{4} c^{3} x^{3}-2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, b^{4} c^{3} x^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, b^{4} c^{3} x^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-3 \sqrt {-\frac {b}{a}}\, a^{4} d^{3} x +3 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{3} b c \,d^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+6 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{3} b c \,d^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {-\frac {b}{a}}\, a^{2} b^{2} c^{2} d x +7 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{2} b^{2} c^{2} d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a^{2} b^{2} c^{2} d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+3 \sqrt {-\frac {b}{a}}\, a \,b^{3} c^{3} x -2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a \,b^{3} c^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a \,b^{3} c^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )}{3 \sqrt {d \,x^{2}+c}\, \left (a d -b c \right )^{3} \sqrt {-\frac {b}{a}}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(5/2)/(d*x^2+c)^(3/2),x)

[Out]

-1/3*(-3*x^5*a^2*b^2*d^3*(-1/a*b)^(1/2)-7*x^5*a*b^3*c*d^2*(-1/a*b)^(1/2)+2*x^5*b^4*c^2*d*(-1/a*b)^(1/2)+3*Elli
pticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^2*a^2*b^2*c*d^2*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)+7*EllipticE(
(-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^2*a*b^3*c^2*d*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)-2*EllipticE((-1/a*b)
^(1/2)*x,(a/b/c*d)^(1/2))*x^2*b^4*c^3*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)+6*EllipticF((-1/a*b)^(1/2)*x,(a/
b/c*d)^(1/2))*x^2*a^2*b^2*c*d^2*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)-8*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)
^(1/2))*x^2*a*b^3*c^2*d*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)+2*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*
x^2*b^4*c^3*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)-6*x^3*a^3*b*d^3*(-1/a*b)^(1/2)-8*x^3*a^2*b^2*c*d^2*(-1/a*b
)^(1/2)-4*x^3*a*b^3*c^2*d*(-1/a*b)^(1/2)+2*x^3*b^4*c^3*(-1/a*b)^(1/2)+3*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(
1/2))*a^3*b*c*d^2*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)+7*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*b^
2*c^2*d*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)-2*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b^3*c^3*((d*x^
2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)+6*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^3*b*c*d^2*((d*x^2+c)/c)^(1/2
)*((b*x^2+a)/a)^(1/2)-8*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*b^2*c^2*d*((d*x^2+c)/c)^(1/2)*((b*x^2+
a)/a)^(1/2)+2*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b^3*c^3*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)-3*
x*a^4*d^3*(-1/a*b)^(1/2)-8*x*a^2*b^2*c^2*d*(-1/a*b)^(1/2)+3*x*a*b^3*c^3*(-1/a*b)^(1/2))/(d*x^2+c)^(1/2)/(a*d-b
*c)^3/(-1/a*b)^(1/2)/a^2/c/(b*x^2+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(d*x^2 + c)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^2+a\right )}^{5/2}\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)^(3/2)),x)

[Out]

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(5/2)/(d*x**2+c)**(3/2),x)

[Out]

Integral(1/((a + b*x**2)**(5/2)*(c + d*x**2)**(3/2)), x)

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