3.1094 \(\int \sqrt{d x} (a+b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=148 \[ \frac{2 a (d x)^{3/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{3}{4};-\frac{3}{2},-\frac{3}{2};\frac{7}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{3 d \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

(2*a*(d*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[3/4, -3/2, -3/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-
2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*d*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sq
rt[b^2 - 4*a*c])])

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Rubi [A]  time = 0.130043, antiderivative size = 148, 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} \[ \frac{2 a (d x)^{3/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{3}{4};-\frac{3}{2},-\frac{3}{2};\frac{7}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{3 d \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(d*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[3/4, -3/2, -3/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-
2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*d*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sq
rt[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 \sqrt{d x} \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac{\left (a \sqrt{a+b x^2+c x^4}\right ) \int \sqrt{d x} \left (1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}\right )^{3/2} \left (1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )^{3/2} \, 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 a (d x)^{3/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{3}{4};-\frac{3}{2},-\frac{3}{2};\frac{7}{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 )}{3 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 = 0.728047, size = 417, normalized size = 2.82 \[ \frac{2 x \sqrt{d x} \left (7 \left (209 a^2 c+12 a b^2+328 a b c x^2+286 a c^2 x^4+131 b^2 c x^4+12 b^3 x^2+196 b c^2 x^6+77 c^3 x^8\right )+12 b x^2 \left (36 a c-5 b^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{7}{4};\frac{1}{2},\frac{1}{2};\frac{11}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )-28 a \left (3 b^2-44 a c\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{3}{4};\frac{1}{2},\frac{1}{2};\frac{7}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )\right )}{8085 c \sqrt{a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*Sqrt[d*x]*(7*(12*a*b^2 + 209*a^2*c + 12*b^3*x^2 + 328*a*b*c*x^2 + 131*b^2*c*x^4 + 286*a*c^2*x^4 + 196*b*c
^2*x^6 + 77*c^3*x^8) - 28*a*(3*b^2 - 44*a*c)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*S
qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sq
rt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 12*b*(-5*b^2 + 36*a*c)*x^2*Sqrt[(b - Sqrt[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[
7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(8085*c*Sqrt[a
+ b*x^2 + c*x^4])

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Maple [F]  time = 0.28, size = 0, normalized size = 0. \begin{align*} \int \sqrt{dx} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError