3.14.29 \(\int \frac {(-1+x) (-1+k x) (3-2 (1+k) x+k x^2)}{x ((1-x) x (1-k x))^{3/4} (-1+(1+k) x-k x^2+d x^3)} \, dx\)

Optimal. Leaf size=96 \[ 2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )+\frac {4 \sqrt [4]{k x^3+(-k-1) x^2+x}}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 16.65, 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+x) (-1+k x) \left (3-2 (1+k) x+k x^2\right )}{x ((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 + x)*(-1 + k*x)*(3 - 2*(1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - k*x^2 + d
*x^3)),x]

[Out]

(4*(1 - x)^(3/4)*(1 - k*x)^(3/4)*AppellF1[-3/4, -1/4, -1/4, 1/4, x, k*x])/((1 - x)*x*(1 - k*x))^(3/4) + (8*k*(
1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[Int][(x^4*(1 - x^4)^(1/4)*(1 - k*x^4)^(1/4))/(1 - (1 +
k)*x^4 + k*x^8 - d*x^12), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(3/4) + (4*(1 + k)*(1 - x)^(3/4)*x^(3/4)*(1
- k*x)^(3/4)*Defer[Subst][Defer[Int][((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4))/(-1 + (1 + k)*x^4 - k*x^8 + d*x^12),
x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(3/4) + (12*d*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[
Int][(x^8*(1 - x^4)^(1/4)*(1 - k*x^4)^(1/4))/(-1 + (1 + k)*x^4 - k*x^8 + d*x^12), x], x, x^(1/4)])/((1 - x)*x*
(1 - k*x))^(3/4)

Rubi steps

\begin {align*} \int \frac {(-1+x) (-1+k x) \left (3-2 (1+k) x+k x^2\right )}{x ((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {(-1+x) (-1+k x) \left (3-2 (1+k) x+k x^2\right )}{(1-x)^{3/4} x^{7/4} (1-k x)^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=-\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{1-x} (-1+k x) \left (3-2 (1+k) x+k x^2\right )}{x^{7/4} (1-k x)^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{1-x} \sqrt [4]{1-k x} \left (3-2 (1+k) x+k x^2\right )}{x^{7/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (3-2 (1+k) x^4+k x^8\right )}{x^4 \left (-1+(1+k) x^4-k x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (-\frac {3 \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{x^4}+\frac {\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1-k+2 k x^4-3 d x^8\right )}{1-(1+k) x^4+k x^8-d x^{12}}\right ) \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1-k+2 k x^4-3 d x^8\right )}{1-(1+k) x^4+k x^8-d x^{12}} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}-\frac {\left (12 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{x^4} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 (1-x)^{3/4} (1-k x)^{3/4} F_1\left (-\frac {3}{4};-\frac {1}{4},-\frac {1}{4};\frac {1}{4};x,k x\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {2 k x^4 \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{1-(1+k) x^4+k x^8-d x^{12}}+\frac {(1+k) \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{-1+(1+k) x^4-k x^8+d x^{12}}+\frac {3 d x^8 \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{-1+(1+k) x^4-k x^8+d x^{12}}\right ) \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 (1-x)^{3/4} (1-k x)^{3/4} F_1\left (-\frac {3}{4};-\frac {1}{4},-\frac {1}{4};\frac {1}{4};x,k x\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (12 d (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8 \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{-1+(1+k) x^4-k x^8+d x^{12}} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (8 k (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{1-(1+k) x^4+k x^8-d x^{12}} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1+k) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4}}{-1+(1+k) x^4-k x^8+d x^{12}} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 1.70, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) (-1+k x) \left (3-2 (1+k) x+k x^2\right )}{x ((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-1 + x)*(-1 + k*x)*(3 - 2*(1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - k*x
^2 + d*x^3)),x]

[Out]

Integrate[((-1 + x)*(-1 + k*x)*(3 - 2*(1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - k*x
^2 + d*x^3)), x]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 4.25, size = 96, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}+2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x)*(-1 + k*x)*(3 - 2*(1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k
)*x - k*x^2 + d*x^3)),x]

[Out]

(4*(x + (-1 - k)*x^2 + k*x^3)^(1/4))/x + 2*d^(1/4)*ArcTan[(d^(1/4)*x)/(x + (-1 - k)*x^2 + k*x^3)^(1/4)] - 2*d^
(1/4)*ArcTanh[(d^(1/4)*x)/(x + (-1 - k)*x^2 + k*x^3)^(1/4)]

________________________________________________________________________________________

fricas [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+x)*(k*x-1)*(3-2*(1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x^2+d*x^3),x, algorithm=
"fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)*(k*x-1)*(3-2*(1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x^2+d*x^3),x, algorithm=
"giac")

[Out]

integrate((k*x^2 - 2*(k + 1)*x + 3)*(k*x - 1)*(x - 1)/((d*x^3 - k*x^2 + (k + 1)*x - 1)*((k*x - 1)*(x - 1)*x)^(
3/4)*x), x)

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+x \right ) \left (k x -1\right ) \left (3-2 \left (1+k \right ) x +k \,x^{2}\right )}{x \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-1+\left (1+k \right ) x -k \,x^{2}+d \,x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+x)*(k*x-1)*(3-2*(1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x^2+d*x^3),x)

[Out]

int((-1+x)*(k*x-1)*(3-2*(1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x^2+d*x^3),x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)*(k*x-1)*(3-2*(1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x^2+d*x^3),x, algorithm=
"maxima")

[Out]

integrate((k*x^2 - 2*(k + 1)*x + 3)*(k*x - 1)*(x - 1)/((d*x^3 - k*x^2 + (k + 1)*x - 1)*((k*x - 1)*(x - 1)*x)^(
3/4)*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (k\,x-1\right )\,\left (x-1\right )\,\left (k\,x^2-2\,x\,\left (k+1\right )+3\right )}{x\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (d\,x^3-k\,x^2+\left (k+1\right )\,x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((k*x - 1)*(x - 1)*(k*x^2 - 2*x*(k + 1) + 3))/(x*(x*(k*x - 1)*(x - 1))^(3/4)*(d*x^3 + x*(k + 1) - k*x^2 -
1)),x)

[Out]

int(((k*x - 1)*(x - 1)*(k*x^2 - 2*x*(k + 1) + 3))/(x*(x*(k*x - 1)*(x - 1))^(3/4)*(d*x^3 + x*(k + 1) - k*x^2 -
1)), x)

________________________________________________________________________________________

sympy [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+x)*(k*x-1)*(3-2*(1+k)*x+k*x**2)/x/((1-x)*x*(-k*x+1))**(3/4)/(-1+(1+k)*x-k*x**2+d*x**3),x)

[Out]

Timed out

________________________________________________________________________________________