3.206 \(\int \frac{1}{(a+b x^2)^{7/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=334 \[ -\frac{\sqrt{c} \sqrt{d} \sqrt{a+b x^2} \left (15 a^2 d^2-11 a b c d+4 b^2 c^2\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{15 a^3 \sqrt{c+d x^2} (b c-a d)^3 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{\sqrt{b} \sqrt{c+d x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{15 a^{5/2} \sqrt{a+b x^2} (b c-a d)^3 \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac{4 b x \sqrt{c+d x^2} (b c-2 a d)}{15 a^2 \left (a+b x^2\right )^{3/2} (b c-a d)^2}+\frac{b x \sqrt{c+d x^2}}{5 a \left (a+b x^2\right )^{5/2} (b c-a d)} \]

[Out]

(b*x*Sqrt[c + d*x^2])/(5*a*(b*c - a*d)*(a + b*x^2)^(5/2)) + (4*b*(b*c - 2*a*d)*x*Sqrt[c + d*x^2])/(15*a^2*(b*c
- a*d)^2*(a + b*x^2)^(3/2)) + (Sqrt[b]*(8*b^2*c^2 - 23*a*b*c*d + 23*a^2*d^2)*Sqrt[c + d*x^2]*EllipticE[ArcTan
[(Sqrt[b]*x)/Sqrt[a]], 1 - (a*d)/(b*c)])/(15*a^(5/2)*(b*c - a*d)^3*Sqrt[a + b*x^2]*Sqrt[(a*(c + d*x^2))/(c*(a
+ b*x^2))]) - (Sqrt[c]*Sqrt[d]*(4*b^2*c^2 - 11*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]
*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*a^3*(b*c - a*d)^3*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.277528, antiderivative size = 334, normalized size of antiderivative = 1., 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} \[ -\frac{\sqrt{c} \sqrt{d} \sqrt{a+b x^2} \left (15 a^2 d^2-11 a b c d+4 b^2 c^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 a^3 \sqrt{c+d x^2} (b c-a d)^3 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{\sqrt{b} \sqrt{c+d x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{15 a^{5/2} \sqrt{a+b x^2} (b c-a d)^3 \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac{4 b x \sqrt{c+d x^2} (b c-2 a d)}{15 a^2 \left (a+b x^2\right )^{3/2} (b c-a d)^2}+\frac{b x \sqrt{c+d x^2}}{5 a \left (a+b x^2\right )^{5/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(b*x*Sqrt[c + d*x^2])/(5*a*(b*c - a*d)*(a + b*x^2)^(5/2)) + (4*b*(b*c - 2*a*d)*x*Sqrt[c + d*x^2])/(15*a^2*(b*c
- a*d)^2*(a + b*x^2)^(3/2)) + (Sqrt[b]*(8*b^2*c^2 - 23*a*b*c*d + 23*a^2*d^2)*Sqrt[c + d*x^2]*EllipticE[ArcTan
[(Sqrt[b]*x)/Sqrt[a]], 1 - (a*d)/(b*c)])/(15*a^(5/2)*(b*c - a*d)^3*Sqrt[a + b*x^2]*Sqrt[(a*(c + d*x^2))/(c*(a
+ b*x^2))]) - (Sqrt[c]*Sqrt[d]*(4*b^2*c^2 - 11*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]
*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*a^3*(b*c - a*d)^3*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

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 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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}} \, dx &=\frac{b x \sqrt{c+d x^2}}{5 a (b c-a d) \left (a+b x^2\right )^{5/2}}-\frac{\int \frac{-4 b c+5 a d-3 b d x^2}{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}} \, dx}{5 a (b c-a d)}\\ &=\frac{b x \sqrt{c+d x^2}}{5 a (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac{4 b (b c-2 a d) x \sqrt{c+d x^2}}{15 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{\int \frac{8 b^2 c^2-19 a b c d+15 a^2 d^2+4 b d (b c-2 a d) x^2}{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}} \, dx}{15 a^2 (b c-a d)^2}\\ &=\frac{b x \sqrt{c+d x^2}}{5 a (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac{4 b (b c-2 a d) x \sqrt{c+d x^2}}{15 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac{\left (d \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 a^2 (b c-a d)^3}+\frac{\left (b \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right )\right ) \int \frac{\sqrt{c+d x^2}}{\left (a+b x^2\right )^{3/2}} \, dx}{15 a^2 (b c-a d)^3}\\ &=\frac{b x \sqrt{c+d x^2}}{5 a (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac{4 b (b c-2 a d) x \sqrt{c+d x^2}}{15 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{\sqrt{b} \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{15 a^{5/2} (b c-a d)^3 \sqrt{a+b x^2} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac{\sqrt{c} \sqrt{d} \left (4 b^2 c^2-11 a b c d+15 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 )}{15 a^3 (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*}

Mathematica [C]  time = 0.620821, size = 301, normalized size = 0.9 \[ \frac{b x \sqrt{\frac{b}{a}} \left (c+d x^2\right ) \left (\left (a+b x^2\right )^2 \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right )+3 a^2 (b c-a d)^2+4 a \left (a+b x^2\right ) (b c-2 a d) (b c-a d)\right )+i \sqrt{\frac{b x^2}{a}+1} \left (a+b x^2\right )^2 \sqrt{\frac{d x^2}{c}+1} \left (\left (-34 a^2 b c d^2+15 a^3 d^3+27 a b^2 c^2 d-8 b^3 c^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+b c \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )}{15 a^3 \sqrt{\frac{b}{a}} \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(b*Sqrt[b/a]*x*(c + d*x^2)*(3*a^2*(b*c - a*d)^2 + 4*a*(b*c - 2*a*d)*(b*c - a*d)*(a + b*x^2) + (8*b^2*c^2 - 23*
a*b*c*d + 23*a^2*d^2)*(a + b*x^2)^2) + I*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(b*c*(8*b^2*c^2
- 23*a*b*c*d + 23*a^2*d^2)*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (-8*b^3*c^3 + 27*a*b^2*c^2*d - 34
*a^2*b*c*d^2 + 15*a^3*d^3)*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(15*a^3*Sqrt[b/a]*(b*c - a*d)^3*(a
+ b*x^2)^(5/2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.031, size = 1607, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

1/15*(-15*x*a^2*b^3*c^3*(-b/a)^(1/2)+15*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^5*d^3*((b*x^2+a)/a)^(1/2)*
((d*x^2+c)/c)^(1/2)-23*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a*b^4*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c
)/c)^(1/2)-34*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a^2*b^3*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1
/2)-46*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^2*b^3*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-68*
EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^3*b^2*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+27*Ellipti
cF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a*b^4*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+46*EllipticE(x*(-b/
a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^3*b^2*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+30*EllipticF(x*(-b/a)^(1/2
),(a*d/b/c)^(1/2))*x^2*a^4*b*d^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-16*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)
^(1/2))*x^2*a*b^4*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+23*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^4
*b*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-23*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*b^2*c^2*d*((
b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-34*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^4*b*c*d^2*((b*x^2+a)/a)^(
1/2)*((d*x^2+c)/c)^(1/2)+27*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*b^2*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^
2+c)/c)^(1/2)-23*x^7*a^2*b^3*d^3*(-b/a)^(1/2)-34*x*a^4*b*c*d^2*(-b/a)^(1/2)-13*x^3*a^3*b^2*c*d^2*(-b/a)^(1/2)+
43*x^3*a^2*b^3*c^2*d*(-b/a)^(1/2)+35*x^5*a^2*b^3*c*d^2*(-b/a)^(1/2)+23*x^7*a*b^4*c*d^2*(-b/a)^(1/2)-8*Elliptic
F(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*b^5*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+8*EllipticE(x*(-b/a)^(1/
2),(a*d/b/c)^(1/2))*a^2*b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-8*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(
1/2))*a^2*b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+15*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a^3
*b^2*d^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+16*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a*b^4*c^3*((
b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+54*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^2*b^3*c^2*d*((b*x^2+a
)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-8*x^7*b^5*c^2*d*(-b/a)^(1/2)-54*x^5*a^3*b^2*d^3*(-b/a)^(1/2)-34*x^3*a^4*b*d^3*(
-b/a)^(1/2)-20*x^3*a*b^4*c^3*(-b/a)^(1/2)-8*x^5*b^5*c^3*(-b/a)^(1/2)+41*x*a^3*b^2*c^2*d*(-b/a)^(1/2)+8*Ellipti
cE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*b^5*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+3*x^5*a*b^4*c^2*d*(-b/a
)^(1/2)+23*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a^2*b^3*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)
)/(d*x^2+c)^(1/2)/(a*d-b*c)^3/(-b/a)^(1/2)/a^3/(b*x^2+a)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{7}{2}} \sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(7/2)*sqrt(d*x^2 + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{b^{4} d x^{10} +{\left (b^{4} c + 4 \, a b^{3} d\right )} x^{8} + 2 \,{\left (2 \, a b^{3} c + 3 \, a^{2} b^{2} d\right )} x^{6} + a^{4} c + 2 \,{\left (3 \, a^{2} b^{2} c + 2 \, a^{3} b d\right )} x^{4} +{\left (4 \, a^{3} b c + a^{4} d\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right )^{\frac{7}{2}} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{7}{2}} \sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(7/2)*sqrt(d*x^2 + c)), x)