3.955 \(\int \frac {(c x)^{3/2}}{\sqrt [4]{a-b x^2}} \, dx\)

Optimal. Leaf size=308 \[ -\frac {a c^{3/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{8 \sqrt {2} b^{5/4}}+\frac {a c^{3/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{8 \sqrt {2} b^{5/4}}-\frac {a c^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {a c^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{4 \sqrt {2} b^{5/4}}-\frac {c \sqrt {c x} \left (a-b x^2\right )^{3/4}}{2 b} \]

[Out]

1/8*a*c^(3/2)*arctan(-1+b^(1/4)*2^(1/2)*(c*x)^(1/2)/(-b*x^2+a)^(1/4)/c^(1/2))/b^(5/4)*2^(1/2)+1/8*a*c^(3/2)*ar
ctan(1+b^(1/4)*2^(1/2)*(c*x)^(1/2)/(-b*x^2+a)^(1/4)/c^(1/2))/b^(5/4)*2^(1/2)-1/16*a*c^(3/2)*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^(5/4)*2^(1/2)+1/16*a*c^(3/2)*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^(5/4)*2^(1/2)-1/2*c*(
-b*x^2+a)^(3/4)*(c*x)^(1/2)/b

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Rubi [A]  time = 0.27, antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {321, 329, 240, 211, 1165, 628, 1162, 617, 204} \[ -\frac {a c^{3/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{8 \sqrt {2} b^{5/4}}+\frac {a c^{3/2} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{8 \sqrt {2} b^{5/4}}-\frac {a c^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {a c^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{4 \sqrt {2} b^{5/4}}-\frac {c \sqrt {c x} \left (a-b x^2\right )^{3/4}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/(a - b*x^2)^(1/4),x]

[Out]

-(c*Sqrt[c*x]*(a - b*x^2)^(3/4))/(2*b) - (a*c^(3/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2
)^(1/4))])/(4*Sqrt[2]*b^(5/4)) + (a*c^(3/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))
])/(4*Sqrt[2]*b^(5/4)) - (a*c^(3/2)*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)])/(8*Sqrt[2]*b^(5/4)) + (a*c^(3/2)*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)])/(8*Sqrt[2]*b^(5/4))

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 211

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

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

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

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Mathematica [A]  time = 0.15, size = 241, normalized size = 0.78 \[ -\frac {(c x)^{3/2} \left (8 \sqrt [4]{b} \sqrt {x} \left (a-b x^2\right )^{3/4}+\sqrt {2} a \log \left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a-b x^2}}+1\right )-\sqrt {2} a \log \left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a-b x^2}}+1\right )+2 \sqrt {2} a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a-b x^2}}\right )-2 \sqrt {2} a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a-b x^2}}+1\right )\right )}{16 b^{5/4} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((c*x)^(3/2)*(8*b^(1/4)*Sqrt[x]*(a - b*x^2)^(3/4) + 2*Sqrt[2]*a*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/(a
- b*x^2)^(1/4)] - 2*Sqrt[2]*a*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)] + Sqrt[2]*a*Log[1 + (Sqr
t[b]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)] - Sqrt[2]*a*Log[1 + (Sqrt[b]*x)/Sqrt[a
- b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)]))/(b^(5/4)*x^(3/2))

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fricas [A]  time = 0.62, size = 340, normalized size = 1.10 \[ -\frac {4 \, {\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c + 4 \, \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \arctan \left (-\frac {\left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {3}{4}} {\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a b^{4} c - {\left (b^{5} x^{2} - a b^{4}\right )} \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {3}{4}} \sqrt {-\frac {\sqrt {-b x^{2} + a} a^{2} c^{3} x - \sqrt {-\frac {a^{4} c^{6}}{b^{5}}} {\left (b^{3} x^{2} - a b^{2}\right )}}{b x^{2} - a}}}{a^{4} b c^{6} x^{2} - a^{5} c^{6}}\right ) + \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c + \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} {\left (b^{2} x^{2} - a b\right )}}{b x^{2} - a}\right ) - \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c - \left (-\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} {\left (b^{2} x^{2} - a b\right )}}{b x^{2} - a}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(-b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

-1/8*(4*(-b*x^2 + a)^(3/4)*sqrt(c*x)*c + 4*(-a^4*c^6/b^5)^(1/4)*b*arctan(-((-a^4*c^6/b^5)^(3/4)*(-b*x^2 + a)^(
3/4)*sqrt(c*x)*a*b^4*c - (b^5*x^2 - a*b^4)*(-a^4*c^6/b^5)^(3/4)*sqrt(-(sqrt(-b*x^2 + a)*a^2*c^3*x - sqrt(-a^4*
c^6/b^5)*(b^3*x^2 - a*b^2))/(b*x^2 - a)))/(a^4*b*c^6*x^2 - a^5*c^6)) + (-a^4*c^6/b^5)^(1/4)*b*log(((-b*x^2 + a
)^(3/4)*sqrt(c*x)*a*c + (-a^4*c^6/b^5)^(1/4)*(b^2*x^2 - a*b))/(b*x^2 - a)) - (-a^4*c^6/b^5)^(1/4)*b*log(((-b*x
^2 + a)^(3/4)*sqrt(c*x)*a*c - (-a^4*c^6/b^5)^(1/4)*(b^2*x^2 - a*b))/(b*x^2 - a)))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(-b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((c*x)^(3/2)/(-b*x^2 + a)^(1/4), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {3}{2}}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)/(-b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(3/2)/(-b*x^2+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(-b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(3/2)/(-b*x^2 + a)^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x\right )}^{3/2}}{{\left (a-b\,x^2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)/(a - b*x^2)^(1/4),x)

[Out]

int((c*x)^(3/2)/(a - b*x^2)^(1/4), x)

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sympy [C]  time = 2.20, size = 46, normalized size = 0.15 \[ \frac {c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(3/2)/(-b*x**2+a)**(1/4),x)

[Out]

c**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), b*x**2*exp_polar(2*I*pi)/a)/(2*a**(1/4)*gamma(9/4))

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