3.22.55 \(\int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{a x+\sqrt {-b+a^2 x^2}} \sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}} \, dx\)

Optimal. Leaf size=157 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt [3]{\sqrt {a^2 x^2-b}+a x}+c}}{\sqrt [4]{c}}\right )}{2 a c^{5/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt [3]{\sqrt {a^2 x^2-b}+a x}+c}}{\sqrt [4]{c}}\right )}{2 a c^{5/4}}-\frac {3 \left (\sqrt [3]{\sqrt {a^2 x^2-b}+a x}+c\right )^{3/4}}{a c \sqrt [3]{\sqrt {a^2 x^2-b}+a x}} \]

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Rubi [F]  time = 1.26, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{a x+\sqrt {-b+a^2 x^2}} \sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[Int][1/(Sqrt[-b + a^2*x^2]*(a*x + Sqrt[-b + a^2*x^2])^(1/3)*(c + (a*x + Sqrt[-b + a^2*x^2])^(1/3))^(1/4)
), x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{a x+\sqrt {-b+a^2 x^2}} \sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}} \, dx &=\int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{a x+\sqrt {-b+a^2 x^2}} \sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}} \, dx\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 74, normalized size = 0.47 \begin {gather*} \frac {4 \left (\sqrt [3]{\sqrt {a^2 x^2-b}+a x}+c\right )^{3/4} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};\frac {c+\sqrt [3]{a x+\sqrt {a^2 x^2-b}}}{c}\right )}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.41, size = 157, normalized size = 1.00 \begin {gather*} -\frac {3 \left (c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}\right )^{3/4}}{a c \sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}-\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}}{\sqrt [4]{c}}\right )}{2 a c^{5/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{c+\sqrt [3]{a x+\sqrt {-b+a^2 x^2}}}}{\sqrt [4]{c}}\right )}{2 a c^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.60, size = 291, normalized size = 1.85 \begin {gather*} \frac {3 \, {\left (4 \, a b c \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a^{2} c^{3} \sqrt {\frac {1}{a^{4} c^{5}}} + \sqrt {c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{3}}}} a c \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {1}{4}} - a {\left (c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{3}}\right )}^{\frac {1}{4}} c \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {1}{4}}\right ) + a b c \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {1}{4}} \log \left (a^{3} c^{4} \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {3}{4}} + {\left (c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{3}}\right )}^{\frac {1}{4}}\right ) - a b c \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {1}{4}} \log \left (-a^{3} c^{4} \left (\frac {1}{a^{4} c^{5}}\right )^{\frac {3}{4}} + {\left (c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{3}}\right )}^{\frac {1}{4}}\right ) - 4 \, {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {2}{3}} {\left (a x - \sqrt {a^{2} x^{2} - b}\right )} {\left (c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{3}}\right )}^{\frac {3}{4}}\right )}}{4 \, a b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {a^{2} x^{2}-b}\, \left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{3}} \left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{3}}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.98, size = 99, normalized size = 0.63 \begin {gather*} -\frac {12\,{\left (\frac {c}{{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/3}}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {5}{4};\ \frac {9}{4};\ -\frac {c}{{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/3}}\right )}{5\,a\,{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/3}\,{\left (c+{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/3}\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{c + \sqrt [3]{a x + \sqrt {a^{2} x^{2} - b}}} \sqrt [3]{a x + \sqrt {a^{2} x^{2} - b}} \sqrt {a^{2} x^{2} - b}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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