3.323 \(\int \frac{(a+b x^2)^{3/4}}{c+d x^2} \, dx\)

Optimal. Leaf size=244 \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \sqrt{a d-b c} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}-\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \sqrt{a d-b c} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}+\frac{2 b x}{d \sqrt [4]{a+b x^2}}-\frac{2 \sqrt{a} \sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{d \sqrt [4]{a+b x^2}} \]

[Out]

(2*b*x)/(d*(a + b*x^2)^(1/4)) - (2*Sqrt[a]*Sqrt[b]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]
/2, 2])/(d*(a + b*x^2)^(1/4)) + (a^(1/4)*Sqrt[-(b*c) + a*d]*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/
Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^(3/2)*x) - (a^(1/4)*Sqrt[-(b*c) + a*d]*Sqrt[-(
(b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^(3/2)*
x)

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Rubi [A]  time = 0.150341, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {402, 229, 227, 196, 399, 490, 1218} \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \sqrt{a d-b c} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}-\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \sqrt{a d-b c} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}+\frac{2 b x}{d \sqrt [4]{a+b x^2}}-\frac{2 \sqrt{a} \sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{d \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/4)/(c + d*x^2),x]

[Out]

(2*b*x)/(d*(a + b*x^2)^(1/4)) - (2*Sqrt[a]*Sqrt[b]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]
/2, 2])/(d*(a + b*x^2)^(1/4)) + (a^(1/4)*Sqrt[-(b*c) + a*d]*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/
Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^(3/2)*x) - (a^(1/4)*Sqrt[-(b*c) + a*d]*Sqrt[-(
(b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^(3/2)*
x)

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{3/4}}{c+d x^2} \, dx &=\frac{b \int \frac{1}{\sqrt [4]{a+b x^2}} \, dx}{d}-\frac{(b c-a d) \int \frac{1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{d}\\ &=-\frac{\left (2 (b c-a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{a}} \left (b c-a d+d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d x}+\frac{\left (b \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx}{d \sqrt [4]{a+b x^2}}\\ &=\frac{2 b x}{d \sqrt [4]{a+b x^2}}+\frac{\left ((b c-a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-b c+a d}-\sqrt{d} x^2\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d^{3/2} x}-\frac{\left ((b c-a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-b c+a d}+\sqrt{d} x^2\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d^{3/2} x}-\frac{\left (b \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{d \sqrt [4]{a+b x^2}}\\ &=\frac{2 b x}{d \sqrt [4]{a+b x^2}}-\frac{2 \sqrt{a} \sqrt{b} \sqrt [4]{1+\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{d \sqrt [4]{a+b x^2}}+\frac{\sqrt [4]{a} \sqrt{-b c+a d} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}-\frac{\sqrt [4]{a} \sqrt{-b c+a d} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^{3/2} x}\\ \end{align*}

Mathematica [C]  time = 0.158206, size = 161, normalized size = 0.66 \[ \frac{6 a c x \left (a+b x^2\right )^{3/4} F_1\left (\frac{1}{2};-\frac{3}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{\left (c+d x^2\right ) \left (x^2 \left (3 b c F_1\left (\frac{3}{2};\frac{1}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-4 a d F_1\left (\frac{3}{2};-\frac{3}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )+6 a c F_1\left (\frac{1}{2};-\frac{3}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^2)^(3/4)/(c + d*x^2),x]

[Out]

(6*a*c*x*(a + b*x^2)^(3/4)*AppellF1[1/2, -3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/((c + d*x^2)*(6*a*c*Appell
F1[1/2, -3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(-4*a*d*AppellF1[3/2, -3/4, 2, 5/2, -((b*x^2)/a), -((d
*x^2)/c)] + 3*b*c*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/4)/(d*x^2+c),x)

[Out]

int((b*x^2+a)^(3/4)/(d*x^2+c),x)

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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}{4}}}{d x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{\frac{3}{4}}}{c + d x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/4)/(d*x**2+c),x)

[Out]

Integral((a + b*x**2)**(3/4)/(c + d*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="giac")

[Out]

Exception raised: TypeError