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

Optimal. Leaf size=343 \[ \frac {3 a^2 c^{5/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}+\frac {3 a^2 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{32 \sqrt {2} b^{7/4}}+\frac {(c x)^{7/2} \sqrt [4]{a-b x^2}}{4 c}-\frac {a c (c x)^{3/2} \sqrt [4]{a-b x^2}}{16 b} \]

[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+3/64*a^2*c^(5/2)*arctan(-1+b^(1/4)
*2^(1/2)*(c*x)^(1/2)/(-b*x^2+a)^(1/4)/c^(1/2))/b^(7/4)*2^(1/2)+3/64*a^2*c^(5/2)*arctan(1+b^(1/4)*2^(1/2)*(c*x)
^(1/2)/(-b*x^2+a)^(1/4)/c^(1/2))/b^(7/4)*2^(1/2)+3/128*a^2*c^(5/2)*ln(c^(1/2)-b^(1/4)*2^(1/2)*(c*x)^(1/2)/(-b*
x^2+a)^(1/4)+x*b^(1/2)*c^(1/2)/(-b*x^2+a)^(1/2))/b^(7/4)*2^(1/2)-3/128*a^2*c^(5/2)*ln(c^(1/2)+b^(1/4)*2^(1/2)*
(c*x)^(1/2)/(-b*x^2+a)^(1/4)+x*b^(1/2)*c^(1/2)/(-b*x^2+a)^(1/2))/b^(7/4)*2^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {279, 321, 329, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {3 a^2 c^{5/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}+\frac {3 a^2 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{32 \sqrt {2} b^{7/4}}+\frac {(c x)^{7/2} \sqrt [4]{a-b x^2}}{4 c}-\frac {a c (c x)^{3/2} \sqrt [4]{a-b x^2}}{16 b} \]

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[1
- (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))])/(32*Sqrt[2]*b^(7/4)) + (3*a^2*c^(5/2)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))])/(32*Sqrt[2]*b^(7/4)) + (3*a^2*c^(5/2)*Log[Sqrt[c] + (
Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)])/(64*Sqrt[2]*b^(7/4)) - (3
*a^2*c^(5/2)*Log[Sqrt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)
])/(64*Sqrt[2]*b^(7/4))

Rule 204

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

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 297

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

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

Rule 1165

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

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\right ) \operatorname {Subst}\left (\int \frac {c-\sqrt {b} x^2}{1+\frac {b x^4}{c^2}} \, dx,x,\frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{32 b^{3/2}}+\frac {\left (3 a^2 c\right ) \operatorname {Subst}\left (\int \frac {c+\sqrt {b} x^2}{1+\frac {b x^4}{c^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 {\left (3 a^2 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {c}}{\sqrt [4]{b}}+2 x}{-\frac {c}{\sqrt {b}}-\frac {\sqrt {2} \sqrt {c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}+\frac {\left (3 a^2 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {c}}{\sqrt [4]{b}}-2 x}{-\frac {c}{\sqrt {b}}+\frac {\sqrt {2} \sqrt {c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}+\frac {\left (3 a^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {c}{\sqrt {b}}-\frac {\sqrt {2} \sqrt {c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 b^2}+\frac {\left (3 a^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {c}{\sqrt {b}}+\frac {\sqrt {2} \sqrt {c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 b^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} \log \left (\sqrt {c}+\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}+\frac {\left (3 a^2 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}-\frac {\left (3 a^2 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}\\ &=-\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 (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}+\frac {3 a^2 c^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{32 \sqrt {2} b^{7/4}}+\frac {3 a^2 c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}-\frac {3 a^2 c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}\right )}{64 \sqrt {2} b^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 88, normalized size = 0.26 \[ \frac {c (c x)^{3/2} \sqrt [4]{a-b x^2} \left (a \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {b x^2}{a}\right )+\sqrt [4]{1-\frac {b x^2}{a}} \left (b x^2-a\right )\right )}{4 b \sqrt [4]{1-\frac {b x^2}{a}}} \]

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 = -0.00 \[ \int {\left (c\,x\right )}^{5/2}\,{\left (a-b\,x^2\right )}^{1/4} \,d x \]

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 = 9.29, size = 48, normalized size = 0.14 \[ \frac {\sqrt [4]{a} c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} \]

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(2*I*pi)/a)/(2*gamma(11/4))

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