3.951 \(\int \frac{x^2 (a+b x^2)^{3/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=335 \[ \frac{2 c^{3/2} \sqrt{a+b x^2} (2 b c-3 a d) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{c} \sqrt{a+b x^2} \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 b d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{x \sqrt{a+b x^2} \left (-\frac{3 a^2 d}{b}+13 a c-\frac{8 b c^2}{d}\right )}{15 d \sqrt{c+d x^2}}-\frac{2 x \sqrt{a+b x^2} \sqrt{c+d x^2} (2 b c-3 a d)}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d} \]

[Out]

-((13*a*c - (8*b*c^2)/d - (3*a^2*d)/b)*x*Sqrt[a + b*x^2])/(15*d*Sqrt[c + d*x^2]) - (2*(2*b*c - 3*a*d)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*d^2) + (b*x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(5*d) - (Sqrt[c]*(8*b^2*c^2 - 13
*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b*d^(5/2)*S
qrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (2*c^(3/2)*(2*b*c - 3*a*d)*Sqrt[a + b*x^2]*EllipticF[A
rcTan[(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
])

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Rubi [A]  time = 0.340289, antiderivative size = 335, normalized size of antiderivative = 1., 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 = {477, 582, 531, 418, 492, 411} \[ -\frac{\sqrt{c} \sqrt{a+b x^2} \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 b d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{x \sqrt{a+b x^2} \left (-\frac{3 a^2 d}{b}+13 a c-\frac{8 b c^2}{d}\right )}{15 d \sqrt{c+d x^2}}+\frac{2 c^{3/2} \sqrt{a+b x^2} (2 b c-3 a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 d^{5/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{2 x \sqrt{a+b x^2} \sqrt{c+d x^2} (2 b c-3 a d)}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((13*a*c - (8*b*c^2)/d - (3*a^2*d)/b)*x*Sqrt[a + b*x^2])/(15*d*Sqrt[c + d*x^2]) - (2*(2*b*c - 3*a*d)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*d^2) + (b*x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(5*d) - (Sqrt[c]*(8*b^2*c^2 - 13
*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b*d^(5/2)*S
qrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (2*c^(3/2)*(2*b*c - 3*a*d)*Sqrt[a + b*x^2]*EllipticF[A
rcTan[(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 477

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

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{x^2 \left (a+b x^2\right )^{3/2}}{\sqrt{c+d x^2}} \, dx &=\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d}+\frac{\int \frac{x^2 \left (-a (3 b c-5 a d)-2 b (2 b c-3 a d) x^2\right )}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{5 d}\\ &=-\frac{2 (2 b c-3 a d) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d}-\frac{\int \frac{-2 a b c (2 b c-3 a d)-b \left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 b d^2}\\ &=-\frac{2 (2 b c-3 a d) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d}+\frac{(2 a c (2 b c-3 a d)) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 d^2}+\frac{\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\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^2-13 a b c d+3 a^2 d^2\right ) x \sqrt{a+b x^2}}{15 b d^2 \sqrt{c+d x^2}}-\frac{2 (2 b c-3 a d) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d}+\frac{2 c^{3/2} (2 b c-3 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 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^2-13 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}{15 b d^2}\\ &=\frac{\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) x \sqrt{a+b x^2}}{15 b d^2 \sqrt{c+d x^2}}-\frac{2 (2 b c-3 a d) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 d^2}+\frac{b x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{5 d}-\frac{\sqrt{c} \left (8 b^2 c^2-13 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 )}{15 b 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{2 c^{3/2} (2 b c-3 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 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 = 0.424786, size = 245, normalized size = 0.73 \[ \frac{i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (9 a^2 d^2-17 a b c d+8 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )-i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (6 a d-4 b c+3 b d x^2\right )}{15 d^3 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(-4*b*c + 6*a*d + 3*b*d*x^2) - I*c*(8*b^2*c^2 - 13*a*b*c*d + 3*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 - 17*
a*b*c*d + 9*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*Sqrt[b/a]*d^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.018, size = 544, normalized size = 1.6 \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}^{2}{d}^{3}-9\,\sqrt{-{\frac{b}{a}}}{x}^{5}ab{d}^{3}+\sqrt{-{\frac{b}{a}}}{x}^{5}{b}^{2}c{d}^{2}-6\,\sqrt{-{\frac{b}{a}}}{x}^{3}{a}^{2}{d}^{3}-5\,\sqrt{-{\frac{b}{a}}}{x}^{3}abc{d}^{2}+4\,\sqrt{-{\frac{b}{a}}}{x}^{3}{b}^{2}{c}^{2}d+9\,\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}c{d}^{2}-17\,\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 ) ab{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}^{2}{c}^{3}-3\,\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}c{d}^{2}+13\,\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 ) ab{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}^{2}{c}^{3}-6\,\sqrt{-{\frac{b}{a}}}x{a}^{2}c{d}^{2}+4\,\sqrt{-{\frac{b}{a}}}xab{c}^{2}d \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+a)^(3/2)/(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^2*d^3-9*(-b/a)^(1/2)*x^5*a*b*d^3+(-b/a)^(1/2)*x^5
*b^2*c*d^2-6*(-b/a)^(1/2)*x^3*a^2*d^3-5*(-b/a)^(1/2)*x^3*a*b*c*d^2+4*(-b/a)^(1/2)*x^3*b^2*c^2*d+9*((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*c*d^2-17*((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*c^2*d+8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Ell
ipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^2*c^3-3*((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*c*d^2+13*((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*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^2*c^3-6*
(-b/a)^(1/2)*x*a^2*c*d^2+4*(-b/a)^(1/2)*x*a*b*c^2*d)/d^3/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}} x^{2}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/2)*x^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{{\left (b x^{4} + a x^{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(x^2*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}} x^{2}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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