3.364 \(\int x^{3/2} (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=350 \[ \frac{28 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{56 b^4 x^{3/2} \left (b+c x^2\right )}{1105 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}-\frac{56 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2} \]

[Out]

(56*b^4*x^(3/2)*(b + c*x^2))/(1105*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (56*b^3*Sqrt[x]*Sqrt[b
*x^2 + c*x^4])/(3315*c^2) + (8*b^2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(663*c) + (12*b*x^(9/2)*Sqrt[b*x^2 + c*x^4])/2
21 + (2*x^(5/2)*(b*x^2 + c*x^4)^(3/2))/17 - (56*b^(17/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S
qrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(1105*c^(11/4)*Sqrt[b*x^2 + c*x^4]) + (28*b^
(17/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/
b^(1/4)], 1/2])/(1105*c^(11/4)*Sqrt[b*x^2 + c*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.433556, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2021, 2024, 2032, 329, 305, 220, 1196} \[ \frac{56 b^4 x^{3/2} \left (b+c x^2\right )}{1105 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{28 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}-\frac{56 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(56*b^4*x^(3/2)*(b + c*x^2))/(1105*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (56*b^3*Sqrt[x]*Sqrt[b
*x^2 + c*x^4])/(3315*c^2) + (8*b^2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(663*c) + (12*b*x^(9/2)*Sqrt[b*x^2 + c*x^4])/2
21 + (2*x^(5/2)*(b*x^2 + c*x^4)^(3/2))/17 - (56*b^(17/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S
qrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(1105*c^(11/4)*Sqrt[b*x^2 + c*x^4]) + (28*b^
(17/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/
b^(1/4)], 1/2])/(1105*c^(11/4)*Sqrt[b*x^2 + c*x^4])

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

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

Rubi steps

\begin{align*} \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{1}{17} (6 b) \int x^{7/2} \sqrt{b x^2+c x^4} \, dx\\ &=\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{1}{221} \left (12 b^2\right ) \int \frac{x^{11/2}}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}-\frac{\left (28 b^3\right ) \int \frac{x^{7/2}}{\sqrt{b x^2+c x^4}} \, dx}{663 c}\\ &=-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{\left (28 b^4\right ) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx}{1105 c^2}\\ &=-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{\left (28 b^4 x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{1105 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{\left (56 b^4 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{1105 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}+\frac{\left (56 b^{9/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{1105 c^{5/2} \sqrt{b x^2+c x^4}}-\frac{\left (56 b^{9/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{b}}}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{1105 c^{5/2} \sqrt{b x^2+c x^4}}\\ &=\frac{56 b^4 x^{3/2} \left (b+c x^2\right )}{1105 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{56 b^3 \sqrt{x} \sqrt{b x^2+c x^4}}{3315 c^2}+\frac{8 b^2 x^{5/2} \sqrt{b x^2+c x^4}}{663 c}+\frac{12}{221} b x^{9/2} \sqrt{b x^2+c x^4}+\frac{2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}-\frac{56 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{28 b^{17/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 c^{11/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0618634, size = 101, normalized size = 0.29 \[ \frac{2 \sqrt{x} \sqrt{x^2 \left (b+c x^2\right )} \left (7 b^3 \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{b}\right )-\left (7 b-13 c x^2\right ) \left (b+c x^2\right )^2 \sqrt{\frac{c x^2}{b}+1}\right )}{221 c^2 \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*(-((7*b - 13*c*x^2)*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]) + 7*b^3*Hypergeometric
2F1[-3/2, 3/4, 7/4, -((c*x^2)/b)]))/(221*c^2*Sqrt[1 + (c*x^2)/b])

________________________________________________________________________________________

Maple [A]  time = 0.198, size = 248, normalized size = 0.7 \begin{align*}{\frac{2}{3315\, \left ( c{x}^{2}+b \right ) ^{2}{c}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 195\,{x}^{10}{c}^{5}+480\,{x}^{8}b{c}^{4}+305\,{x}^{6}{b}^{2}{c}^{3}+84\,{b}^{5}\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) -42\,{b}^{5}\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) -8\,{x}^{4}{b}^{3}{c}^{2}-28\,{x}^{2}{b}^{4}c \right ){x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3315*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c^3*(195*x^10*c^5+480*x^8*b*c^4+305*x^6*b^2*c^3+84*b^5*((c*x+(-
b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*Ell
ipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-42*b^5*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(
1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(
1/2))^(1/2),1/2*2^(1/2))-8*x^4*b^3*c^2-28*x^2*b^4*c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^5 + b*x^3)*sqrt(c*x^4 + b*x^2)*sqrt(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(x**(3/2)*(c*x**4+b*x**2)**(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}\right )}^{\frac{3}{2}} x^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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