3.200 \(\int \frac{(a+b x^2)^{5/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=344 \[ \frac{\sqrt{c} \sqrt{a+b x^2} \left (15 a^2 d^2-11 a b c d+4 b^2 c^2\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{x \sqrt{a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right )}{15 d^2 \sqrt{c+d x^2}}-\frac{\sqrt{c} \sqrt{a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{4 b x \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-2 a d)}{15 d^2}+\frac{b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.282602, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {416, 528, 531, 418, 492, 411} \[ \frac{x \sqrt{a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right )}{15 d^2 \sqrt{c+d x^2}}+\frac{\sqrt{c} \sqrt{a+b x^2} \left (15 a^2 d^2-11 a b c d+4 b^2 c^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{c} \sqrt{a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{4 b x \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-2 a d)}{15 d^2}+\frac{b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 416

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 528

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

Mathematica [C]  time = 0.478191, size = 260, normalized size = 0.76 \[ \frac{-i \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (-34 a^2 b c d^2+15 a^3 d^3+27 a b^2 c^2 d-8 b^3 c^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )-i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )+b d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (11 a d-4 b c+3 b d x^2\right )}{15 d^3 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(-4*b*c + 11*a*d + 3*b*d*x^2) - I*b*c*(8*b^2*c^2 - 23*a*b*c*d + 23*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*(-8*b^3*c^3
+ 27*a*b^2*c^2*d - 34*a^2*b*c*d^2 + 15*a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sq
rt[b/a]*x], (a*d)/(b*c)])/(15*Sqrt[b/a]*d^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 615, normalized size = 1.8 \begin{align*}{\frac{1}{15\,{d}^{3} \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) }\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 3\,\sqrt{-{\frac{b}{a}}}{x}^{7}{b}^{3}{d}^{3}+14\,\sqrt{-{\frac{b}{a}}}{x}^{5}a{b}^{2}{d}^{3}-\sqrt{-{\frac{b}{a}}}{x}^{5}{b}^{3}c{d}^{2}+11\,\sqrt{-{\frac{b}{a}}}{x}^{3}{a}^{2}b{d}^{3}+10\,\sqrt{-{\frac{b}{a}}}{x}^{3}a{b}^{2}c{d}^{2}-4\,\sqrt{-{\frac{b}{a}}}{x}^{3}{b}^{3}{c}^{2}d+15\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){a}^{3}{d}^{3}-34\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){a}^{2}bc{d}^{2}+27\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) a{b}^{2}{c}^{2}d-8\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){b}^{3}{c}^{3}+23\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){a}^{2}bc{d}^{2}-23\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) a{b}^{2}{c}^{2}d+8\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){b}^{3}{c}^{3}+11\,\sqrt{-{\frac{b}{a}}}x{a}^{2}bc{d}^{2}-4\,\sqrt{-{\frac{b}{a}}}xa{b}^{2}{c}^{2}d \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(3*(-b/a)^(1/2)*x^7*b^3*d^3+14*(-b/a)^(1/2)*x^5*a*b^2*d^3-(-b/a)^(1/2)*x^
5*b^3*c*d^2+11*(-b/a)^(1/2)*x^3*a^2*b*d^3+10*(-b/a)^(1/2)*x^3*a*b^2*c*d^2-4*(-b/a)^(1/2)*x^3*b^3*c^2*d+15*((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*d^3-34*((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*d^2+27*((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^2*d-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Elliptic
F(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^3+23*((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*d^2-23*((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^2*d+8*((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^3+11
*(-b/a)^(1/2)*x*a^2*b*c*d^2-4*(-b/a)^(1/2)*x*a*b^2*c^2*d)/d^3/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/(-b/a)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{5}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{5}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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