Integrand size = 19, antiderivative size = 326 \[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\frac {1}{2} d e x^2 \sqrt {a+c x^4}+\frac {2 a e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{15} x \left (5 d^2+3 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {a d e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}-\frac {2 a^{5/4} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{3/4} \left (5 \sqrt {c} d^2+3 \sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}} \] Output:
1/2*d*e*x^2*(c*x^4+a)^(1/2)+2/5*a*e^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+c ^(1/2)*x^2)+1/15*x*(3*e^2*x^2+5*d^2)*(c*x^4+a)^(1/2)+1/2*a*d*e*arctanh(c^( 1/2)*x^2/(c*x^4+a)^(1/2))/c^(1/2)-2/5*a^(5/4)*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)+1/15*a^(3/4)*(5*c^(1/2)*d^2+ 3*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*arctan(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/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 = 10.16 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.45 \[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\frac {\sqrt {a+c x^4} \left (6 \sqrt {c} d^2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^4}{a}\right )+e \left (3 d \left (\sqrt {c} x^2 \sqrt {1+\frac {c x^4}{a}}+\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )\right )+2 \sqrt {c} e x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )\right )\right )}{6 \sqrt {c} \sqrt {1+\frac {c x^4}{a}}} \] Input:
Integrate[(d + e*x)^2*Sqrt[a + c*x^4],x]
Output:
(Sqrt[a + c*x^4]*(6*Sqrt[c]*d^2*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x ^4)/a)] + e*(3*d*(Sqrt[c]*x^2*Sqrt[1 + (c*x^4)/a] + Sqrt[a]*ArcSinh[(Sqrt[ c]*x^2)/Sqrt[a]]) + 2*Sqrt[c]*e*x^3*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c *x^4)/a)])))/(6*Sqrt[c]*Sqrt[1 + (c*x^4)/a])
Time = 0.72 (sec) , antiderivative size = 326, 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.
| \(\displaystyle \int \sqrt {a+c x^4} (d+e x)^2 \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\sqrt {a+c x^4} \left (d^2+e^2 x^2\right )+2 d e x \sqrt {a+c x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (3 \sqrt {a} e^2+5 \sqrt {c} d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {2 a^{5/4} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a d e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {1}{15} x \sqrt {a+c x^4} \left (5 d^2+3 e^2 x^2\right )+\frac {1}{2} d e x^2 \sqrt {a+c x^4}+\frac {2 a e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}\) |
Input:
Int[(d + e*x)^2*Sqrt[a + c*x^4],x]
Output:
(d*e*x^2*Sqrt[a + c*x^4])/2 + (2*a*e^2*x*Sqrt[a + c*x^4])/(5*Sqrt[c]*(Sqrt [a] + Sqrt[c]*x^2)) + (x*(5*d^2 + 3*e^2*x^2)*Sqrt[a + c*x^4])/15 + (a*d*e* ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (2*a^(5/4)*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])/(5*c^(3/4)*Sqrt[a + c*x^4]) + (a^(3/4) *(5*Sqrt[c]*d^2 + 3*Sqrt[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])/ (15*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 = 1.06 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(\frac {x \left (6 e^{2} x^{2}+15 d e x +10 d^{2}\right ) \sqrt {c \,x^{4}+a}}{30}+\frac {a \left (\frac {10 d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {6 i e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {15 d e \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}\right )}{15}\) | \(234\) |
| default | \(d^{2} \left (\frac {x \sqrt {c \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+e^{2} \left (\frac {x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\right )+2 d e \left (\frac {x^{2} \sqrt {c \,x^{4}+a}}{4}+\frac {a \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{4 \sqrt {c}}\right )\) | \(248\) |
| elliptic | \(\frac {e^{2} x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {d e \,x^{2} \sqrt {c \,x^{4}+a}}{2}+\frac {d^{2} x \sqrt {c \,x^{4}+a}}{3}+\frac {2 a \,d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {a d e \ln \left (2 \sqrt {c}\, x^{2}+2 \sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}+\frac {2 i a^{\frac {3}{2}} e^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(252\) |
Input:
int((e*x+d)^2*(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/30*x*(6*e^2*x^2+15*d*e*x+10*d^2)*(c*x^4+a)^(1/2)+1/15*a*(10*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)+6*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))+15/2*d*e*ln(c^(1/ 2)*x^2+(c*x^4+a)^(1/2))/c^(1/2))
Time = 0.16 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.51 \[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\frac {24 \, a \sqrt {c} e^{2} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 15 \, a \sqrt {c} d e x \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 8 \, {\left (5 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (6 \, c e^{2} x^{4} + 15 \, c d e x^{3} + 10 \, c d^{2} x^{2} + 12 \, a e^{2}\right )} \sqrt {c x^{4} + a}}{60 \, c x} \] Input:
integrate((e*x+d)^2*(c*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/60*(24*a*sqrt(c)*e^2*x*(-a/c)^(3/4)*elliptic_e(arcsin((-a/c)^(1/4)/x), - 1) + 15*a*sqrt(c)*d*e*x*log(-2*c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 8*(5*c*d^2 - 3*a*e^2)*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f(arcsin((-a/c)^(1 /4)/x), -1) + 2*(6*c*e^2*x^4 + 15*c*d*e*x^3 + 10*c*d^2*x^2 + 12*a*e^2)*sqr t(c*x^4 + a))/(c*x)
Result contains complex when optimal does not.
Time = 2.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.42 \[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\frac {\sqrt {a} d^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} d e x^{2} \sqrt {1 + \frac {c x^{4}}{a}}}{2} + \frac {\sqrt {a} e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a d e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {c}} \] Input:
integrate((e*x+d)**2*(c*x**4+a)**(1/2),x)
Output:
sqrt(a)*d**2*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**4*exp_polar(I*pi )/a)/(4*gamma(5/4)) + sqrt(a)*d*e*x**2*sqrt(1 + c*x**4/a)/2 + sqrt(a)*e**2 *x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*g amma(7/4)) + a*d*e*asinh(sqrt(c)*x**2/sqrt(a))/(2*sqrt(c))
\[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\int { \sqrt {c x^{4} + a} {\left (e x + d\right )}^{2} \,d x } \] Input:
integrate((e*x+d)^2*(c*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(c*x^4 + a)*(e*x + d)^2, x)
\[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\int { \sqrt {c x^{4} + a} {\left (e x + d\right )}^{2} \,d x } \] Input:
integrate((e*x+d)^2*(c*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(c*x^4 + a)*(e*x + d)^2, x)
Timed out. \[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\int \sqrt {c\,x^4+a}\,{\left (d+e\,x\right )}^2 \,d x \] Input:
int((a + c*x^4)^(1/2)*(d + e*x)^2,x)
Output:
int((a + c*x^4)^(1/2)*(d + e*x)^2, x)
\[ \int (d+e x)^2 \sqrt {a+c x^4} \, dx=\frac {20 \sqrt {c \,x^{4}+a}\, c \,d^{2} x +30 \sqrt {c \,x^{4}+a}\, c d e \,x^{2}+12 \sqrt {c \,x^{4}+a}\, c \,e^{2} x^{3}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{4}+a}-\sqrt {c}\, x^{2}\right ) a d e +15 \sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}\right ) a d e +40 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{4}+a}d x \right ) a c \,d^{2}+24 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) a c \,e^{2}}{60 c} \] Input:
int((e*x+d)^2*(c*x^4+a)^(1/2),x)
Output:
(20*sqrt(a + c*x**4)*c*d**2*x + 30*sqrt(a + c*x**4)*c*d*e*x**2 + 12*sqrt(a + c*x**4)*c*e**2*x**3 - 15*sqrt(c)*log(sqrt(a + c*x**4) - sqrt(c)*x**2)*a *d*e + 15*sqrt(c)*log(sqrt(a + c*x**4) + sqrt(c)*x**2)*a*d*e + 40*int(sqrt (a + c*x**4)/(a + c*x**4),x)*a*c*d**2 + 24*int((sqrt(a + c*x**4)*x**2)/(a + c*x**4),x)*a*c*e**2)/(60*c)