3.2.72 \(\int (a+b x^2)^{3/2} (c+d x^2)^{3/2} \, dx\) [172]

Optimal. Leaf size=410 \[ -\frac {2 (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right ) x \sqrt {a+b x^2}}{35 b^2 d \sqrt {c+d x^2}}+\frac {1}{35} \left (9 a c+\frac {b c^2}{d}-\frac {2 a^2 d}{b}\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {2 \sqrt {c} (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b^2 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {c^{3/2} \left (b^2 c^2-18 a b c d+a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

-2/35*(a*d+b*c)*(a^2*d^2-6*a*b*c*d+b^2*c^2)*x*(b*x^2+a)^(1/2)/b^2/d/(d*x^2+c)^(1/2)-1/35*c^(3/2)*(a^2*d^2-18*a
*b*c*d+b^2*c^2)*(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/d^(3/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+2/35*(a*d+b*c)*(a^2*d^2-6
*a*b*c*d+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^2/d^(3/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+2/35*(-a*d+4*
b*c)*x*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/b+1/7*d*x*(b*x^2+a)^(5/2)*(d*x^2+c)^(1/2)/b+1/35*(9*a*c+b*c^2/d-2*a^2*d
/b)*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 410, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {427, 542, 545, 429, 506, 422} \begin {gather*} \frac {2 \sqrt {c} \sqrt {a+b x^2} (a d+b c) \left (a^2 d^2-6 a b c d+b^2 c^2\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b^2 d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {c^{3/2} \sqrt {a+b x^2} \left (a^2 d^2-18 a b c d+b^2 c^2\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {2 x \sqrt {a+b x^2} (a d+b c) \left (a^2 d^2-6 a b c d+b^2 c^2\right )}{35 b^2 d \sqrt {c+d x^2}}+\frac {1}{35} x \sqrt {a+b x^2} \sqrt {c+d x^2} \left (-\frac {2 a^2 d}{b}+9 a c+\frac {b c^2}{d}\right )+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {2 x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (4 b c-a d)}{35 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c + a*d)*(b^2*c^2 - 6*a*b*c*d + a^2*d^2)*x*Sqrt[a + b*x^2])/(35*b^2*d*Sqrt[c + d*x^2]) + ((9*a*c + (b*c
^2)/d - (2*a^2*d)/b)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/35 + (2*(4*b*c - a*d)*x*(a + b*x^2)^(3/2)*Sqrt[c + d*x
^2])/(35*b) + (d*x*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(7*b) + (2*Sqrt[c]*(b*c + a*d)*(b^2*c^2 - 6*a*b*c*d + a^
2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b^2*d^(3/2)*Sqrt[(c*(a + b
*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (c^(3/2)*(b^2*c^2 - 18*a*b*c*d + a^2*d^2)*Sqrt[a + b*x^2]*EllipticF
[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d
*x^2])

Rule 422

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

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 429

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

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 542

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

Rule 545

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]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^{3/2} \, dx &=\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {\int \frac {\left (a+b x^2\right )^{3/2} \left (c (7 b c-a d)+2 d (4 b c-a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{7 b}\\ &=\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {\int \frac {\sqrt {a+b x^2} \left (3 a c d (9 b c-a d)+3 d \left (b^2 c^2+9 a b c d-2 a^2 d^2\right ) x^2\right )}{\sqrt {c+d x^2}} \, dx}{35 b d}\\ &=\frac {1}{35} \left (9 a c+\frac {b c^2}{d}-\frac {2 a^2 d}{b}\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {\int \frac {-3 a c d \left (b^2 c^2-18 a b c d+a^2 d^2\right )-6 d (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{105 b d^2}\\ &=\frac {1}{35} \left (9 a c+\frac {b c^2}{d}-\frac {2 a^2 d}{b}\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}-\frac {\left (a c \left (b^2 c^2-18 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{35 b d}-\frac {\left (2 (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right )\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{35 b d}\\ &=-\frac {2 (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right ) x \sqrt {a+b x^2}}{35 b^2 d \sqrt {c+d x^2}}+\frac {1}{35} \left (9 a c+\frac {b c^2}{d}-\frac {2 a^2 d}{b}\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}-\frac {c^{3/2} \left (b^2 c^2-18 a b c d+a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\left (2 c (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{35 b^2 d}\\ &=-\frac {2 (b c+a d) \left (b^2 c^2-6 a b c d+a^2 d^2\right ) x \sqrt {a+b x^2}}{35 b^2 d \sqrt {c+d x^2}}+\frac {1}{35} \left (9 a c+\frac {b c^2}{d}-\frac {2 a^2 d}{b}\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 (4 b c-a d) x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{35 b}+\frac {d x \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{7 b}+\frac {2 \sqrt {c} (b c+a d) \left (b^2 c^2-6 a b c d+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 )}{35 b^2 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {c^{3/2} \left (b^2 c^2-18 a b c d+a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{35 b d^{3/2} \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] Result contains complex when optimal does not.
time = 4.39, size = 302, normalized size = 0.74 \begin {gather*} \frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (a^2 d^2+a b d \left (17 c+8 d x^2\right )+b^2 \left (c^2+8 c d x^2+5 d^2 x^4\right )\right )+2 i c \left (b^3 c^3-5 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \left (2 b^3 c^3-11 a b^2 c^2 d+8 a^2 b c d^2+a^3 d^3\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{35 b \sqrt {\frac {b}{a}} d^2 \sqrt {a+b x^2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(a^2*d^2 + a*b*d*(17*c + 8*d*x^2) + b^2*(c^2 + 8*c*d*x^2 + 5*d^2*x^4))
+ (2*I)*c*(b^3*c^3 - 5*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*Elliptic
E[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*c*(2*b^3*c^3 - 11*a*b^2*c^2*d + 8*a^2*b*c*d^2 + a^3*d^3)*Sqrt[1 + (
b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(35*b*Sqrt[b/a]*d^2*Sqrt[a + b*x
^2]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.10, size = 780, normalized size = 1.90 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(5*(-b/a)^(1/2)*b^3*d^4*x^9+13*(-b/a)^(1/2)*a*b^2*d^4*x^7+13*(-b/a)^(1/2)
*b^3*c*d^3*x^7+9*(-b/a)^(1/2)*a^2*b*d^4*x^5+38*(-b/a)^(1/2)*a*b^2*c*d^3*x^5+9*(-b/a)^(1/2)*b^3*c^2*d^2*x^5+(-b
/a)^(1/2)*a^3*d^4*x^3+26*(-b/a)^(1/2)*a^2*b*c*d^3*x^3+26*(-b/a)^(1/2)*a*b^2*c^2*d^2*x^3+(-b/a)^(1/2)*b^3*c^3*d
*x^3+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*c*d^3+8*((b*x^2+a)/
a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^2*b*c^2*d^2-11*((b*x^2+a)/a)^(1/2)*((
d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*b^2*c^3*d+2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1
/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^4-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-
b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*c*d^3+10*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d
/b/c)^(1/2))*a^2*b*c^2*d^2+10*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2)
)*a*b^2*c^3*d-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^4+(-b/
a)^(1/2)*a^3*c*d^3*x+17*(-b/a)^(1/2)*a^2*b*c^2*d^2*x+(-b/a)^(1/2)*a*b^2*c^3*d*x)/b/d^2/(b*d*x^4+a*d*x^2+b*c*x^
2+a*c)/(-b/a)^(1/2)

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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)^(3/2)*(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b x^{2}\right )^{\frac {3}{2}} \left (c + d x^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,x^2+a\right )}^{3/2}\,{\left (d\,x^2+c\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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