Integrand size = 22, antiderivative size = 292 \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\frac {x \sqrt {d+e x^2}}{8 a \left (a-c x^4\right )^2}+\frac {x \sqrt {d+e x^2} \left (7 c d^2-6 a e^2-c d e x^2\right )}{32 a^2 \left (c d^2-a e^2\right ) \left (a-c x^4\right )}+\frac {\left (21 c d^2-34 \sqrt {a} \sqrt {c} d e+12 a e^2\right ) \arctan \left (\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{64 a^{11/4} \sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )^{3/2}}+\frac {\left (21 c d^2+34 \sqrt {a} \sqrt {c} d e+12 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {c} d+\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{64 a^{11/4} \sqrt {c} \left (\sqrt {c} d+\sqrt {a} e\right )^{3/2}} \] Output:
1/8*x*(e*x^2+d)^(1/2)/a/(-c*x^4+a)^2+1/32*x*(e*x^2+d)^(1/2)*(-c*d*e*x^2-6* a*e^2+7*c*d^2)/a^2/(-a*e^2+c*d^2)/(-c*x^4+a)+1/64*(21*c*d^2-34*a^(1/2)*c^( 1/2)*d*e+12*a*e^2)*arctan((c^(1/2)*d-a^(1/2)*e)^(1/2)*x/a^(1/4)/(e*x^2+d)^ (1/2))/a^(11/4)/c^(1/2)/(c^(1/2)*d-a^(1/2)*e)^(3/2)+1/64*(21*c*d^2+34*a^(1 /2)*c^(1/2)*d*e+12*a*e^2)*arctanh((c^(1/2)*d+a^(1/2)*e)^(1/2)*x/a^(1/4)/(e *x^2+d)^(1/2))/a^(11/4)/c^(1/2)/(c^(1/2)*d+a^(1/2)*e)^(3/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.45 (sec) , antiderivative size = 1324, normalized size of antiderivative = 4.53 \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx =\text {Too large to display} \] Input:
Integrate[Sqrt[d + e*x^2]/(a - c*x^4)^3,x]
Output:
((2*x*Sqrt[d + e*x^2]*(10*a^2*e^2 + c^2*d*x^4*(7*d - e*x^2) + a*c*(-11*d^2 + d*e*x^2 - 6*e^2*x^4)))/(a - c*x^4)^2 - (64*a*e^(7/2)*RootSum[c*d^4 - 4* c*d^3*#1 + 6*c*d^2*#1^2 - 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (349*c^3 *d^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 1324*a*c^2*d^4* e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 64*a^2*c*d^2*e^4 *Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 1024*a^3*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 262*c^3*d^5*Log[d + 2*e*x^ 2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 96*a*c^2*d^3*e^2*Log[d + 2*e*x^ 2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 256*a^2*c*d*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 69*c^3*d^4*Log[d + 2*e*x^2 - 2*S qrt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2 - 20*a*c^2*d^2*e^2*Log[d + 2*e*x^2 - 2 *Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2 - 64*a^2*c*e^4*Log[d + 2*e*x^2 - 2*S qrt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2)/(c*d^3 - 3*c*d^2*#1 + 8*a*e^2*#1 + 3* c*d*#1^2 - c*#1^3) & ])/(c^3*d^4) + (e^(3/2)*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^2*#1^2 - 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (13*c^4*d^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 22324*a*c^3*d^6*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 84736*a^2*c^2*d^4*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 4096*a^3*c*d^2*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 65536*a^4*e^8*Log[d + 2*e* x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 58*c^4*d^7*Log[d + 2*e*x^2 - ...
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3}dx\) |
Input:
Int[Sqrt[d + e*x^2]/(a - c*x^4)^3,x]
Output:
$Aborted
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
Time = 0.72 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.22
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \left (\left (-\frac {11 a \,e^{2}}{6}+\frac {7 c \,d^{2}}{4}\right ) \sqrt {d^{2} a c}+a e \left (a \,e^{2}-\frac {13 c \,d^{2}}{12}\right )\right ) \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, d \left (-c \,x^{4}+a \right )^{2} \arctan \left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}}\right )}{16}+\frac {5 \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, \left (\frac {3 \left (\left (\frac {11 a \,e^{2}}{6}-\frac {7 c \,d^{2}}{4}\right ) \sqrt {d^{2} a c}+a e \left (a \,e^{2}-\frac {13 c \,d^{2}}{12}\right )\right ) d \left (-c \,x^{4}+a \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}}\right )}{5}+x \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \sqrt {e \,x^{2}+d}\, \left (a^{2} e^{2}-\frac {11 \left (\frac {6}{11} e^{2} x^{4}-\frac {1}{11} d e \,x^{2}+d^{2}\right ) c a}{10}+\frac {7 x^{4} d \left (-\frac {e \,x^{2}}{7}+d \right ) c^{2}}{10}\right ) \sqrt {d^{2} a c}\right )}{16}}{a^{2} \left (a \,e^{2}-c \,d^{2}\right ) \left (-c \,x^{4}+a \right )^{2} \sqrt {d^{2} a c}\, \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}}\) | \(355\) |
default | \(\text {Expression too large to display}\) | \(9780\) |
Input:
int((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x,method=_RETURNVERBOSE)
Output:
5/16/(d^2*a*c)^(1/2)*(3/5*((-11/6*a*e^2+7/4*c*d^2)*(d^2*a*c)^(1/2)+a*e*(a* e^2-13/12*c*d^2))*((a*e+(d^2*a*c)^(1/2))*a)^(1/2)*d*(-c*x^4+a)^2*arctan((e *x^2+d)^(1/2)/x*a/((-a*e+(d^2*a*c)^(1/2))*a)^(1/2))+((-a*e+(d^2*a*c)^(1/2) )*a)^(1/2)*(3/5*((11/6*a*e^2-7/4*c*d^2)*(d^2*a*c)^(1/2)+a*e*(a*e^2-13/12*c *d^2))*d*(-c*x^4+a)^2*arctanh((e*x^2+d)^(1/2)/x*a/((a*e+(d^2*a*c)^(1/2))*a )^(1/2))+x*((a*e+(d^2*a*c)^(1/2))*a)^(1/2)*(e*x^2+d)^(1/2)*(a^2*e^2-11/10* (6/11*e^2*x^4-1/11*d*e*x^2+d^2)*c*a+7/10*x^4*d*(-1/7*e*x^2+d)*c^2)*(d^2*a* c)^(1/2)))/((-a*e+(d^2*a*c)^(1/2))*a)^(1/2)/((a*e+(d^2*a*c)^(1/2))*a)^(1/2 )/a^2/(a*e^2-c*d^2)/(-c*x^4+a)^2
Leaf count of result is larger than twice the leaf count of optimal. 6046 vs. \(2 (233) = 466\).
Time = 117.34 (sec) , antiderivative size = 6046, normalized size of antiderivative = 20.71 \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x**2+d)**(1/2)/(-c*x**4+a)**3,x)
Output:
Timed out
\[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\int { -\frac {\sqrt {e x^{2} + d}}{{\left (c x^{4} - a\right )}^{3}} \,d x } \] Input:
integrate((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x, algorithm="maxima")
Output:
-integrate(sqrt(e*x^2 + d)/(c*x^4 - a)^3, x)
Timed out. \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\int \frac {\sqrt {e\,x^2+d}}{{\left (a-c\,x^4\right )}^3} \,d x \] Input:
int((d + e*x^2)^(1/2)/(a - c*x^4)^3,x)
Output:
int((d + e*x^2)^(1/2)/(a - c*x^4)^3, x)
\[ \int \frac {\sqrt {d+e x^2}}{\left (a-c x^4\right )^3} \, dx=\int \frac {\sqrt {e \,x^{2}+d}}{\left (-c \,x^{4}+a \right )^{3}}d x \] Input:
int((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x)
Output:
int((e*x^2+d)^(1/2)/(-c*x^4+a)^3,x)