3.1110 \(\int (d x)^m (a+b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=158 \[ \frac{a (d x)^{m+1} \sqrt{a+b x^2+c x^4} F_1\left (\frac{m+1}{2};-\frac{3}{2},-\frac{3}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

(a*(d*x)^(1 + m)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[(1 + m)/2, -3/2, -3/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 -
4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(d*(1 + m)*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 = 0.145507, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1141, 510} \[ \frac{a (d x)^{m+1} \sqrt{a+b x^2+c x^4} F_1\left (\frac{m+1}{2};-\frac{3}{2},-\frac{3}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(d*x)^(1 + m)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[(1 + m)/2, -3/2, -3/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 -
4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(d*(1 + m)*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)^m \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 (d x)^m \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{a (d x)^{1+m} \sqrt{a+b x^2+c x^4} F_1\left (\frac{1+m}{2};-\frac{3}{2},-\frac{3}{2};\frac{3+m}{2};-\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 )}{d (1+m) \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 = 0.277901, size = 357, normalized size = 2.26 \[ \frac{x (d x)^m \sqrt{a+b x^2+c x^4} \left (a \left (m^2+8 m+15\right ) F_1\left (\frac{m+1}{2};-\frac{1}{2},-\frac{1}{2};\frac{m+3}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+(m+1) x^2 \left (c (m+3) x^2 F_1\left (\frac{m+5}{2};-\frac{1}{2},-\frac{1}{2};\frac{m+7}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+b (m+5) F_1\left (\frac{m+3}{2};-\frac{1}{2},-\frac{1}{2};\frac{m+5}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )\right )\right )}{(m+1) (m+3) (m+5) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(d*x)^m*Sqrt[a + b*x^2 + c*x^4]*(a*(15 + 8*m + m^2)*AppellF1[(1 + m)/2, -1/2, -1/2, (3 + m)/2, (-2*c*x^2)/(
b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + (1 + m)*x^2*(b*(5 + m)*AppellF1[(3 + m)/2, -1/2,
-1/2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + c*(3 + m)*x^2*Appe
llF1[(5 + m)/2, -1/2, -1/2, (7 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]
)))/((1 + m)*(3 + m)*(5 + m)*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])])

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m} \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)^m*(c*x^4+b*x^2+a)^(3/2),x)

[Out]

int((d*x)^m*(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}} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)*(d*x)^m, 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}} \left (d x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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