3.597 \(\int (c x)^{7/2} (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=212 \[ \frac{4 a^{15/4} c^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a+b x^2}}-\frac{8 a^3 c^3 \sqrt{c x} \sqrt{a+b x^2}}{231 b^2}+\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c} \]

[Out]

(-8*a^3*c^3*Sqrt[c*x]*Sqrt[a + b*x^2])/(231*b^2) + (8*a^2*c*(c*x)^(5/2)*Sqrt[a + b*x^2])/(385*b) + (4*a*(c*x)^
(9/2)*Sqrt[a + b*x^2])/(55*c) + (2*(c*x)^(9/2)*(a + b*x^2)^(3/2))/(15*c) + (4*a^(15/4)*c^(7/2)*(Sqrt[a] + Sqrt
[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/
2])/(231*b^(9/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.130183, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {279, 321, 329, 220} \[ -\frac{8 a^3 c^3 \sqrt{c x} \sqrt{a+b x^2}}{231 b^2}+\frac{4 a^{15/4} c^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a+b x^2}}+\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)*(a + b*x^2)^(3/2),x]

[Out]

(-8*a^3*c^3*Sqrt[c*x]*Sqrt[a + b*x^2])/(231*b^2) + (8*a^2*c*(c*x)^(5/2)*Sqrt[a + b*x^2])/(385*b) + (4*a*(c*x)^
(9/2)*Sqrt[a + b*x^2])/(55*c) + (2*(c*x)^(9/2)*(a + b*x^2)^(3/2))/(15*c) + (4*a^(15/4)*c^(7/2)*(Sqrt[a] + Sqrt
[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/
2])/(231*b^(9/4)*Sqrt[a + b*x^2])

Rule 279

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

Rule 321

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

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 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]

Rubi steps

\begin{align*} \int (c x)^{7/2} \left (a+b x^2\right )^{3/2} \, dx &=\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{1}{5} (2 a) \int (c x)^{7/2} \sqrt{a+b x^2} \, dx\\ &=\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{1}{55} \left (4 a^2\right ) \int \frac{(c x)^{7/2}}{\sqrt{a+b x^2}} \, dx\\ &=\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}-\frac{\left (4 a^3 c^2\right ) \int \frac{(c x)^{3/2}}{\sqrt{a+b x^2}} \, dx}{77 b}\\ &=-\frac{8 a^3 c^3 \sqrt{c x} \sqrt{a+b x^2}}{231 b^2}+\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{\left (4 a^4 c^4\right ) \int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx}{231 b^2}\\ &=-\frac{8 a^3 c^3 \sqrt{c x} \sqrt{a+b x^2}}{231 b^2}+\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{\left (8 a^4 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{231 b^2}\\ &=-\frac{8 a^3 c^3 \sqrt{c x} \sqrt{a+b x^2}}{231 b^2}+\frac{8 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}{385 b}+\frac{4 a (c x)^{9/2} \sqrt{a+b x^2}}{55 c}+\frac{2 (c x)^{9/2} \left (a+b x^2\right )^{3/2}}{15 c}+\frac{4 a^{15/4} c^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0772995, size = 102, normalized size = 0.48 \[ \frac{2 c^3 \sqrt{c x} \sqrt{a+b x^2} \left (5 a^3 \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )-\left (5 a-11 b x^2\right ) \left (a+b x^2\right )^2 \sqrt{\frac{b x^2}{a}+1}\right )}{165 b^2 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)*(a + b*x^2)^(3/2),x]

[Out]

(2*c^3*Sqrt[c*x]*Sqrt[a + b*x^2]*(-((5*a - 11*b*x^2)*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a]) + 5*a^3*Hypergeometric
2F1[-3/2, 1/4, 5/4, -((b*x^2)/a)]))/(165*b^2*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.019, size = 163, normalized size = 0.8 \begin{align*}{\frac{2\,{c}^{3}}{1155\,{b}^{3}x}\sqrt{cx} \left ( 77\,{b}^{5}{x}^{9}+196\,a{b}^{4}{x}^{7}+10\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{a}^{4}+131\,{a}^{2}{b}^{3}{x}^{5}-8\,{a}^{3}{b}^{2}{x}^{3}-20\,{a}^{4}bx \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)*(b*x^2+a)^(3/2),x)

[Out]

2/1155/x*c^3*(c*x)^(1/2)/(b*x^2+a)^(1/2)*(77*b^5*x^9+196*a*b^4*x^7+10*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*
2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b
)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a^4+131*a^2*b^3*x^5-8*a^3*b^2*x^3-20*a^4*b*x)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)*(c*x)^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral((b*c^3*x^5 + a*c^3*x^3)*sqrt(b*x^2 + a)*sqrt(c*x), x)

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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((c*x)**(7/2)*(b*x**2+a)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)*(c*x)^(7/2), x)