Integrand size = 19, antiderivative size = 264 \[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\frac {e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {d e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{\sqrt {c}}-\frac {\sqrt [4]{a} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}} \] Output:
e^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+c^(1/2)*x^2)+d*e*arctanh(c^(1/2)*x^ 2/(c*x^4+a)^(1/2))/c^(1/2)-a^(1/4)*e^2*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+a)/(a ^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/ 2*2^(1/2))/c^(3/4)/(c*x^4+a)^(1/2)+1/2*(c^(1/2)*d^2+a^(1/2)*e^2)*(a^(1/2)+ c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*a rctan(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(1/4)/c^(3/4)/(c*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.50 \[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\frac {d e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{\sqrt {c}}+\frac {d^2 x \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}}+\frac {e^2 x^3 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )}{3 \sqrt {a+c x^4}} \] Input:
Integrate[(d + e*x)^2/Sqrt[a + c*x^4],x]
Output:
(d*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/Sqrt[c] + (d^2*x*Sqrt[1 + (c* x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)])/Sqrt[a + c*x^4] + (e^2*x^3*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a) ])/(3*Sqrt[a + c*x^4])
Time = 0.59 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2424, 2009}
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.
| \(\displaystyle \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\frac {d^2+e^2 x^2}{\sqrt {a+c x^4}}+\frac {2 d e x}{\sqrt {a+c x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {d e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{\sqrt {c}}+\frac {e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}\) |
Input:
Int[(d + e*x)^2/Sqrt[a + c*x^4],x]
Output:
(e^2*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (d*e*ArcTanh[( Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/Sqrt[c] - (a^(1/4)*e^2*(Sqrt[a] + Sqrt[c]*x ^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4 )*x)/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[a + c*x^4]) + (a^(1/4)*((Sqrt[c]*d^2)/S qrt[a] + e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]* x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[a + c*x^4])
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2 *((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && !PolyQ[Pq, x^(n/2)]
Result contains complex when optimal does not.
Time = 0.74 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {d e \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{\sqrt {c}}\) | \(197\) |
| elliptic | \(\frac {d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {d e \ln \left (2 \sqrt {c}\, x^{2}+2 \sqrt {c \,x^{4}+a}\right )}{\sqrt {c}}+\frac {i e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(200\) |
Input:
int((e*x+d)^2/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
d^2/(I*c^(1/2)/a^(1/2))^(1/2)*(1-I*c^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*c^(1/2) *x^2/a^(1/2))^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I*c^(1/2)/a^(1/2))^(1/2), I)+I*e^2*a^(1/2)/(I*c^(1/2)/a^(1/2))^(1/2)*(1-I*c^(1/2)*x^2/a^(1/2))^(1/2) *(1+I*c^(1/2)*x^2/a^(1/2))^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(x*(I*c ^(1/2)/a^(1/2))^(1/2),I)-EllipticE(x*(I*c^(1/2)/a^(1/2))^(1/2),I))+d*e*ln( c^(1/2)*x^2+(c*x^4+a)^(1/2))/c^(1/2)
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\frac {2 \, a \sqrt {c} e^{2} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + a \sqrt {c} d e x \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, \sqrt {c x^{4} + a} a e^{2}}{2 \, a c x} \] Input:
integrate((e*x+d)^2/(c*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/2*(2*a*sqrt(c)*e^2*x*(-a/c)^(3/4)*elliptic_e(arcsin((-a/c)^(1/4)/x), -1) + a*sqrt(c)*d*e*x*log(-2*c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 2*( c*d^2 - a*e^2)*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f(arcsin((-a/c)^(1/4)/x), - 1) + 2*sqrt(c*x^4 + a)*a*e^2)/(a*c*x)
Result contains complex when optimal does not.
Time = 1.99 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.40 \[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\frac {d e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{\sqrt {c}} + \frac {d^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((e*x+d)**2/(c*x**4+a)**(1/2),x)
Output:
d*e*asinh(sqrt(c)*x**2/sqrt(a))/sqrt(c) + d**2*x*gamma(1/4)*hyper((1/4, 1/ 2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + e**2*x**3*g amma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*g amma(7/4))
\[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{\sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((e*x+d)^2/(c*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x + d)^2/sqrt(c*x^4 + a), x)
\[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{\sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((e*x+d)^2/(c*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate((e*x + d)^2/sqrt(c*x^4 + a), x)
Timed out. \[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{\sqrt {c\,x^4+a}} \,d x \] Input:
int((d + e*x)^2/(a + c*x^4)^(1/2),x)
Output:
int((d + e*x)^2/(a + c*x^4)^(1/2), x)
\[ \int \frac {(d+e x)^2}{\sqrt {a+c x^4}} \, dx=\frac {-\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{4}+a}-\sqrt {c}\, x^{2}\right ) d e +\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}\right ) d e +2 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{4}+a}d x \right ) c \,d^{2}+2 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) c \,e^{2}}{2 c} \] Input:
int((e*x+d)^2/(c*x^4+a)^(1/2),x)
Output:
( - sqrt(c)*log(sqrt(a + c*x**4) - sqrt(c)*x**2)*d*e + sqrt(c)*log(sqrt(a + c*x**4) + sqrt(c)*x**2)*d*e + 2*int(sqrt(a + c*x**4)/(a + c*x**4),x)*c*d **2 + 2*int((sqrt(a + c*x**4)*x**2)/(a + c*x**4),x)*c*e**2)/(2*c)