3.919 \(\int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx\)

Optimal. Leaf size=152 \[ -\frac{a^{5/2} c^2 (c x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}}-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b} \]

[Out]

-(a^2*c^3*Sqrt[c*x]*(a + b*x^2)^(1/4))/(12*b^2) + (a*c*(c*x)^(5/2)*(a + b*x^2)^(1/4))/(30*b) + ((c*x)^(9/2)*(a
+ b*x^2)^(1/4))/(5*c) - (a^(5/2)*c^2*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/
2, 2])/(12*b^(3/2)*(a + b*x^2)^(3/4))

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Rubi [A]  time = 0.107769, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {279, 321, 329, 237, 335, 275, 231} \[ -\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}-\frac{a^{5/2} c^2 (c x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]

[Out]

-(a^2*c^3*Sqrt[c*x]*(a + b*x^2)^(1/4))/(12*b^2) + (a*c*(c*x)^(5/2)*(a + b*x^2)^(1/4))/(30*b) + ((c*x)^(9/2)*(a
+ b*x^2)^(1/4))/(5*c) - (a^(5/2)*c^2*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/
2, 2])/(12*b^(3/2)*(a + b*x^2)^(3/4))

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 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

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

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx &=\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{1}{10} a \int \frac{(c x)^{7/2}}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac{\left (a^2 c^2\right ) \int \frac{(c x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{12 b}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{\left (a^3 c^4\right ) \int \frac{1}{\sqrt{c x} \left (a+b x^2\right )^{3/4}} \, dx}{24 b^2}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{\left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+\frac{b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt{c x}\right )}{12 b^2}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac{\left (a^3 c^3 \left (1+\frac{a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a c^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt{c x}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac{\left (a^3 c^3 \left (1+\frac{a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+\frac{a c^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac{1}{\sqrt{c x}}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac{\left (a^3 c^3 \left (1+\frac{a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a c^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac{1}{c x}\right )}{24 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac{a^2 c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac{a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac{(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac{a^{5/2} c^2 \left (1+\frac{a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0653525, size = 102, normalized size = 0.67 \[ \frac{c^3 \sqrt{c x} \sqrt [4]{a+b x^2} \left (\sqrt [4]{\frac{b x^2}{a}+1} \left (-5 a^2+a b x^2+6 b^2 x^4\right )+5 a^2 \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )\right )}{30 b^2 \sqrt [4]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]

[Out]

(c^3*Sqrt[c*x]*(a + b*x^2)^(1/4)*((1 + (b*x^2)/a)^(1/4)*(-5*a^2 + a*b*x^2 + 6*b^2*x^4) + 5*a^2*Hypergeometric2
F1[-1/4, 1/4, 5/4, -((b*x^2)/a)]))/(30*b^2*(1 + (b*x^2)/a)^(1/4))

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int \left ( cx \right ) ^{{\frac{7}{2}}}\sqrt [4]{b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{4}} \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)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)*(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 x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x} c^{3} x^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)*c^3*x^3, 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)**(1/4),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{1}{4}} \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)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(7/2), x)