3.925 \(\int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx\)

Optimal. Leaf size=147 \[ \frac{3 a^2 c^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}-\frac{3 a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}+\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b} \]

[Out]

(a*c*(c*x)^(3/2)*(a + b*x^2)^(1/4))/(16*b) + ((c*x)^(7/2)*(a + b*x^2)^(1/4))/(4*c) + (3*a^2*c^(5/2)*ArcTan[(b^
(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(32*b^(7/4)) - (3*a^2*c^(5/2)*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt
[c]*(a + b*x^2)^(1/4))])/(32*b^(7/4))

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Rubi [A]  time = 0.0921969, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {279, 321, 329, 331, 298, 205, 208} \[ \frac{3 a^2 c^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}-\frac{3 a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}+\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)*(a + b*x^2)^(1/4),x]

[Out]

(a*c*(c*x)^(3/2)*(a + b*x^2)^(1/4))/(16*b) + ((c*x)^(7/2)*(a + b*x^2)^(1/4))/(4*c) + (3*a^2*c^(5/2)*ArcTan[(b^
(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(32*b^(7/4)) - (3*a^2*c^(5/2)*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt
[c]*(a + b*x^2)^(1/4))])/(32*b^(7/4))

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 331

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int (c x)^{5/2} \sqrt [4]{a+b x^2} \, dx &=\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}+\frac{1}{8} a \int \frac{(c x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}-\frac{\left (3 a^2 c^2\right ) \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{3/4}} \, dx}{32 b}\\ &=\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}-\frac{\left (3 a^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+\frac{b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt{c x}\right )}{16 b}\\ &=\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}-\frac{\left (3 a^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{16 b}\\ &=\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}-\frac{\left (3 a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{c-\sqrt{b} x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{3/2}}+\frac{\left (3 a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{c+\sqrt{b} x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{3/2}}\\ &=\frac{a c (c x)^{3/2} \sqrt [4]{a+b x^2}}{16 b}+\frac{(c x)^{7/2} \sqrt [4]{a+b x^2}}{4 c}+\frac{3 a^2 c^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}-\frac{3 a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{32 b^{7/4}}\\ \end{align*}

Mathematica [C]  time = 0.0442451, size = 85, normalized size = 0.58 \[ \frac{c (c x)^{3/2} \sqrt [4]{a+b x^2} \left (\left (a+b x^2\right ) \sqrt [4]{\frac{b x^2}{a}+1}-a \, _2F_1\left (-\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )\right )}{4 b \sqrt [4]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(5/2)*(a + b*x^2)^(1/4),x]

[Out]

(c*(c*x)^(3/2)*(a + b*x^2)^(1/4)*((a + b*x^2)*(1 + (b*x^2)/a)^(1/4) - a*Hypergeometric2F1[-1/4, 3/4, 7/4, -((b
*x^2)/a)]))/(4*b*(1 + (b*x^2)/a)^(1/4))

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int \left ( cx \right ) ^{{\frac{5}{2}}}\sqrt [4]{b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)*(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(5/2)*(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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(5/2), x)

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Fricas [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)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C]  time = 43.4649, size = 46, normalized size = 0.31 \begin{align*} \frac{\sqrt [4]{a} c^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/2)*(b*x**2+a)**(1/4),x)

[Out]

a**(1/4)*c**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(11/4))

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Giac [B]  time = 3.05243, size = 570, normalized size = 3.88 \begin{align*} -\frac{1}{128} \, a^{2} c^{6}{\left (\frac{6 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} + \frac{2 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \sqrt{{\left | c \right |}}}{\sqrt{c x}}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}}}\right )}{b^{2} c^{4}} + \frac{6 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} - \frac{2 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \sqrt{{\left | c \right |}}}{\sqrt{c x}}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}}}\right )}{b^{2} c^{4}} + \frac{3 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \log \left (\frac{\sqrt{2}{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}{\left | c \right |}}{\sqrt{c x}} + \sqrt{-b}{\left | c \right |} + \frac{\sqrt{b c^{2} x^{2} + a c^{2}}{\left | c \right |}}{c x}\right )}{b^{2} c^{4}} - \frac{3 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \log \left (-\frac{\sqrt{2}{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}{\left | c \right |}}{\sqrt{c x}} + \sqrt{-b}{\left | c \right |} + \frac{\sqrt{b c^{2} x^{2} + a c^{2}}{\left | c \right |}}{c x}\right )}{b^{2} c^{4}} - \frac{8 \,{\left (\frac{3 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} b c^{2} \sqrt{{\left | c \right |}}}{\sqrt{c x}} + \frac{{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}}{\left (b c^{2} + \frac{a c^{2}}{x^{2}}\right )} \sqrt{{\left | c \right |}}}{\sqrt{c x}}\right )} x^{4}}{a^{2} b c^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)*(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

-1/128*a^2*c^6*(6*sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4)*sqrt(abs(c)) + 2*(b*c
^2*x^2 + a*c^2)^(1/4)*sqrt(abs(c))/sqrt(c*x))/((-b)^(1/4)*sqrt(abs(c))))/(b^2*c^4) + 6*sqrt(2)*(-b)^(1/4)*sqrt
(abs(c))*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4)*sqrt(abs(c)) - 2*(b*c^2*x^2 + a*c^2)^(1/4)*sqrt(abs(c))/sqrt(
c*x))/((-b)^(1/4)*sqrt(abs(c))))/(b^2*c^4) + 3*sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*log(sqrt(2)*(b*c^2*x^2 + a*c^2)
^(1/4)*(-b)^(1/4)*abs(c)/sqrt(c*x) + sqrt(-b)*abs(c) + sqrt(b*c^2*x^2 + a*c^2)*abs(c)/(c*x))/(b^2*c^4) - 3*sqr
t(2)*(-b)^(1/4)*sqrt(abs(c))*log(-sqrt(2)*(b*c^2*x^2 + a*c^2)^(1/4)*(-b)^(1/4)*abs(c)/sqrt(c*x) + sqrt(-b)*abs
(c) + sqrt(b*c^2*x^2 + a*c^2)*abs(c)/(c*x))/(b^2*c^4) - 8*(3*(b*c^2*x^2 + a*c^2)^(1/4)*b*c^2*sqrt(abs(c))/sqrt
(c*x) + (b*c^2*x^2 + a*c^2)^(1/4)*(b*c^2 + a*c^2/x^2)*sqrt(abs(c))/sqrt(c*x))*x^4/(a^2*b*c^6))