3.975 \(\int \frac {x^6}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=342 \[ \frac {x \sqrt {a+b x^2} \left (8 a^2 d^2+7 a b c d+8 b^2 c^2\right )}{15 b^3 d^2 \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} \left (8 a^2 d^2+7 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^3 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {4 c^{3/2} \sqrt {a+b x^2} (a d+b c) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {4 x \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d} \]

[Out]

1/15*(8*a^2*d^2+7*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^3/d^2/(d*x^2+c)^(1/2)+4/15*c^(3/2)*(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))*(b*x^2+a)^(1/
2)/b^2/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/15*(8*a^2*d^2+7*a*b*c*d+8*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))*c^(1/2)*(b*x^2+
a)^(1/2)/b^3/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-4/15*(a*d+b*c)*x*(b*x^2+a)^(1/2)*(d*x^2+c
)^(1/2)/b^2/d^2+1/5*x^3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d

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Rubi [A]  time = 0.31, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {479, 582, 531, 418, 492, 411} \[ \frac {x \sqrt {a+b x^2} \left (8 a^2 d^2+7 a b c d+8 b^2 c^2\right )}{15 b^3 d^2 \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} \left (8 a^2 d^2+7 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^3 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {4 c^{3/2} \sqrt {a+b x^2} (a d+b c) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {4 x \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

((8*b^2*c^2 + 7*a*b*c*d + 8*a^2*d^2)*x*Sqrt[a + b*x^2])/(15*b^3*d^2*Sqrt[c + d*x^2]) - (4*(b*c + a*d)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*b^2*d^2) + (x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(5*b*d) - (Sqrt[c]*(8*b^2*c^2
+ 7*a*b*c*d + 8*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b^3*d^(5
/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (4*c^(3/2)*(b*c + a*d)*Sqrt[a + b*x^2]*EllipticF[
ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b^2*d^(5/2)*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 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 479

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

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

Rule 582

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

Rubi steps

\begin {align*} \int \frac {x^6}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx &=\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d}-\frac {\int \frac {x^2 \left (3 a c+4 (b c+a d) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 b d}\\ &=-\frac {4 (b c+a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d}+\frac {\int \frac {4 a c (b c+a d)+\left (8 b^2 c^2+7 a b c d+8 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b^2 d^2}\\ &=-\frac {4 (b c+a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d}+\frac {(4 a c (b c+a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b^2 d^2}+\frac {\left (8 b^2 c^2+7 a b c d+8 a^2 d^2\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b^2 d^2}\\ &=\frac {\left (8 b^2 c^2+7 a b c d+8 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 b^3 d^2 \sqrt {c+d x^2}}-\frac {4 (b c+a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d}+\frac {4 c^{3/2} (b c+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 )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {\left (c \left (8 b^2 c^2+7 a b c d+8 a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 b^3 d^2}\\ &=\frac {\left (8 b^2 c^2+7 a b c d+8 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 b^3 d^2 \sqrt {c+d x^2}}-\frac {4 (b c+a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b^2 d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 b d}-\frac {\sqrt {c} \left (8 b^2 c^2+7 a b c d+8 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 )}{15 b^3 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {4 c^{3/2} (b c+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 )}{15 b^2 d^{5/2} \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 = 0.51, size = 249, normalized size = 0.73 \[ \frac {i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (4 a^2 d^2+3 a b c d+8 b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (8 a^2 d^2+7 a b c d+8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+d x \left (-\sqrt {\frac {b}{a}}\right ) \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 a d+4 b c-3 b d x^2\right )}{15 a^2 d^3 \left (\frac {b}{a}\right )^{5/2} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

(-(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(4*b*c + 4*a*d - 3*b*d*x^2)) - I*c*(8*b^2*c^2 + 7*a*b*c*d + 8*a^2*d^2
)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + I*c*(8*b^2*c^2 + 3*
a*b*c*d + 4*a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(
15*a^2*(b/a)^(5/2)*d^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c} x^{6}}{b d x^{4} + {\left (b c + a d\right )} x^{2} + a c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)), x)

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maple [A]  time = 0.04, size = 546, normalized size = 1.60 \[ -\frac {\left (-3 \sqrt {-\frac {b}{a}}\, b^{2} d^{3} x^{7}+\sqrt {-\frac {b}{a}}\, a b \,d^{3} x^{5}+\sqrt {-\frac {b}{a}}\, b^{2} c \,d^{2} x^{5}+4 \sqrt {-\frac {b}{a}}\, a^{2} d^{3} x^{3}+5 \sqrt {-\frac {b}{a}}\, a b c \,d^{2} x^{3}+4 \sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}+4 \sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x -8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+4 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+4 \sqrt {-\frac {b}{a}}\, a b \,c^{2} d x -7 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+3 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{15 \sqrt {-\frac {b}{a}}\, \left (x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c \right ) b^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/15*(-3*(-1/a*b)^(1/2)*b^2*d^3*x^7+(-1/a*b)^(1/2)*a*b*d^3*x^5+(-1/a*b)^(1/2)*b^2*c*d^2*x^5+4*(-1/a*b)^(1/2)*
a^2*d^3*x^3+5*(-1/a*b)^(1/2)*a*b*c*d^2*x^3+4*(-1/a*b)^(1/2)*b^2*c^2*d*x^3+4*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^
(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*c*d^2+3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Elliptic
F((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b*c^2*d+8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/
2)*x,(a/b/c*d)^(1/2))*b^2*c^3-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(
1/2))*a^2*c*d^2-7*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b*c^2*
d-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b^2*c^3+4*(-1/a*b)^(1/
2)*a^2*c*d^2*x+4*(-1/a*b)^(1/2)*a*b*c^2*d*x)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3/b^2/(-1/a*b)^(1/2)/(b*d*x^4+a
*d*x^2+b*c*x^2+a*c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6}{\sqrt {b\,x^2+a}\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

int(x^6/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**6/(sqrt(a + b*x**2)*sqrt(c + d*x**2)), x)

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