3.208 \(\int (f x)^{3/2} (d+e x^2) (a+b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=299 \[ \frac{2 a d (f x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{3}{2},-\frac{3}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 f \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}+\frac{2 a e (f x)^{9/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{9}{4};-\frac{3}{2},-\frac{3}{2};\frac{13}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{9 f^3 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

(2*a*d*(f*x)^(5/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[5/4, -3/2, -3/2, 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*f*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])]) + (2*a*e*(f*x)^(9/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[9/4, -3/2, -3/2, 13/4, (-2*c*x^2)/(
b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(9*f^3*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])])

________________________________________________________________________________________

Rubi [A]  time = 0.347235, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {1335, 1141, 510} \[ \frac{2 a d (f x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{3}{2},-\frac{3}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 f \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}+\frac{2 a e (f x)^{9/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{9}{4};-\frac{3}{2},-\frac{3}{2};\frac{13}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{9 f^3 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*d*(f*x)^(5/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[5/4, -3/2, -3/2, 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*f*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])]) + (2*a*e*(f*x)^(9/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[9/4, -3/2, -3/2, 13/4, (-2*c*x^2)/(
b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(9*f^3*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 1335

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

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 (f x)^{3/2} \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\int \left (d (f x)^{3/2} \left (a+b x^2+c x^4\right )^{3/2}+\frac{e (f x)^{7/2} \left (a+b x^2+c x^4\right )^{3/2}}{f^2}\right ) \, dx\\ &=d \int (f x)^{3/2} \left (a+b x^2+c x^4\right )^{3/2} \, dx+\frac{e \int (f x)^{7/2} \left (a+b x^2+c x^4\right )^{3/2} \, dx}{f^2}\\ &=\frac{\left (a d \sqrt{a+b x^2+c x^4}\right ) \int (f x)^{3/2} \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{\left (a e \sqrt{a+b x^2+c x^4}\right ) \int (f x)^{7/2} \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}{f^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}}}}\\ &=\frac{2 a d (f x)^{5/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{5}{4};-\frac{3}{2},-\frac{3}{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 f \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 e (f x)^{9/2} \sqrt{a+b x^2+c x^4} F_1\left (\frac{9}{4};-\frac{3}{2},-\frac{3}{2};\frac{13}{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 )}{9 f^3 \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 [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{{\frac{3}{2}}} \left ( e{x}^{2}+d \right ) \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((f*x)^(3/2)*(e*x^2+d)*(c*x^4+b*x^2+a)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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 (e x^{2} + d\right )} \left (f x\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c e f x^{7} +{\left (c d + b e\right )} f x^{5} +{\left (b d + a e\right )} f x^{3} + a d f x\right )} \sqrt{c x^{4} + b x^{2} + a} \sqrt{f x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*e*f*x^7 + (c*d + b*e)*f*x^5 + (b*d + a*e)*f*x^3 + a*d*f*x)*sqrt(c*x^4 + b*x^2 + a)*sqrt(f*x), x)

________________________________________________________________________________________

Sympy [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((f*x)**(3/2)*(e*x**2+d)*(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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 (e x^{2} + d\right )} \left (f x\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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