3.1089 \(\int (d x)^{3/2} \sqrt{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 (d x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{1}{2},-\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 d \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

(2*(d*x)^(5/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[5/4, -1/2, -1/2, 9/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*
c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(5*d*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt
[b^2 - 4*a*c])])

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Rubi [A]  time = 0.207493, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1141, 510} \[ \frac{2 (d x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{1}{2},-\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 d \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(5/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[5/4, -1/2, -1/2, 9/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*
c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(5*d*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt
[b^2 - 4*a*c])])

Rule 1141

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/((1 + (2*c*x^2)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^2)/(b - Rt[b^2 - 4*a*c,
2]))^FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
]))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int (d x)^{3/2} \sqrt{a+b x^2+c x^4} \, dx &=\frac{\sqrt{a+b x^2+c x^4} \int (d x)^{3/2} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}} \, dx}{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}\\ &=\frac{2 (d x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{1}{2},-\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 d \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}\\ \end{align*}

Mathematica [B]  time = 0.584016, size = 365, normalized size = 2.48 \[ \frac{2 d \sqrt{d x} \left (2 x^2 \left (10 a c-3 b^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{5}{4};\frac{1}{2},\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )-10 a b \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{1}{4};\frac{1}{2},\frac{1}{2};\frac{5}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+5 \left (2 b+5 c x^2\right ) \left (a+b x^2+c x^4\right )\right )}{225 c \sqrt{a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*d*Sqrt[d*x]*(5*(2*b + 5*c*x^2)*(a + b*x^2 + c*x^4) - 10*a*b*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqr
t[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/4, 1/2, 1/2, 5/4,
(-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 2*(-3*b^2 + 10*a*c)*x^2*Sqrt[(b - Sq
rt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a
*c])]*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(
225*c*Sqrt[a + b*x^2 + c*x^4])

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{{\frac{3}{2}}}\sqrt{c{x}^{4}+b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*(c*x^4+b*x^2+a)^(1/2),x)

[Out]

int((d*x)^(3/2)*(c*x^4+b*x^2+a)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(3/2)*(c*x**4+b*x**2+a)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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