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]

1/16*a*c*(c*x)^(3/2)*(b*x^2+a)^(1/4)/b+1/4*(c*x)^(7/2)*(b*x^2+a)^(1/4)/c+3/32*a^2*c^(5/2)*arctan(b^(1/4)*(c*x)
^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(7/4)-3/32*a^2*c^(5/2)*arctanh(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/
b^(7/4)

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Rubi [A]  time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.05, 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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {5}{2}}\,{d x} \]

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]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (c x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {5}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)*(a + b*x^2)^(1/4),x)

[Out]

int((c*x)^(5/2)*(a + b*x^2)^(1/4), x)

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sympy [C]  time = 9.47, size = 46, normalized size = 0.31 \[ \frac {\sqrt [4]{a} c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} \]

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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