Integrand size = 19, antiderivative size = 295 \[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\frac {e^3 \sqrt {a+c x^4}}{2 c}+\frac {3 d e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {3 d^2 e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}-\frac {3 \sqrt [4]{a} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {d \left (\sqrt {c} d^2+3 \sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}} \]
3/2*d^2*e*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(1/2)+1/2*e^3*(c*x^4+a)^( 1/2)/c+3*d*e^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-3*a^(1/4)*d *e^2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^( 1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^ 2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^4+a)^(1/ 2)+1/2*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x /a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(3*e^2* a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(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 = 10.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.53 \[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\frac {e^3 \sqrt {a+c x^4}}{2 c}+\frac {3 d^2 e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {d^3 x \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}}+\frac {d e^2 x^3 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}} \]
(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^ 4]])/(2*Sqrt[c]) + (d^3*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)])/Sqrt[a + c*x^4] + (d*e^2*x^3*Sqrt[1 + (c*x^4)/a]*Hyper geometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)])/Sqrt[a + c*x^4]
Time = 0.42 (sec) , antiderivative size = 295, 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.
| \(\displaystyle \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\frac {d^3+3 d e^2 x^2}{\sqrt {a+c x^4}}+\frac {x \left (3 d^2 e+e^3 x^2\right )}{\sqrt {a+c x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}}-\frac {3 \sqrt [4]{a} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {3 d^2 e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {3 d e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \sqrt {a+c x^4}}{2 c}\) |
(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d*e^2*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a ] + Sqrt[c]*x^2)) + (3*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sq rt[c]) - (3*a^(1/4)*d*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]) + (d*(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])/(2*a^(1/4)*c^(3/4)*Sqrt[a + c*x^4])
3.3.12.3.1 Defintions of rubi rules used
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 = 1.60 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(\frac {d^{3} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {e^{3} \sqrt {c \,x^{4}+a}}{2 c}+\frac {3 i d \,e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {3 d^{2} e \ln \left (x^{2} \sqrt {c}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}\) | \(218\) |
| risch | \(\frac {e^{3} \sqrt {c \,x^{4}+a}}{2 c}+d \left (\frac {d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {3 i e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {3 e d \ln \left (x^{2} \sqrt {c}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}\right )\) | \(218\) |
| elliptic | \(\frac {e^{3} \sqrt {c \,x^{4}+a}}{2 c}+\frac {d^{3} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {3 d^{2} e \ln \left (2 x^{2} \sqrt {c}+2 \sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}+\frac {3 i d \,e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(221\) |
d^3/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2) *c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2), I)+1/2*e^3*(c*x^4+a)^(1/2)/c+3*I*d*e^2*a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*( 1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^( 1/2)/c^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1 /2)*c^(1/2))^(1/2),I))+3/2*d^2*e*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))/c^(1/2)
Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\frac {12 \, a \sqrt {c} d e^{2} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, a \sqrt {c} d^{2} e x \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 4 \, {\left (c d^{3} - 3 \, a d e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (a e^{3} x + 6 \, a d e^{2}\right )} \sqrt {c x^{4} + a}}{4 \, a c x} \]
1/4*(12*a*sqrt(c)*d*e^2*x*(-a/c)^(3/4)*elliptic_e(arcsin((-a/c)^(1/4)/x), -1) + 3*a*sqrt(c)*d^2*e*x*log(-2*c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a ) + 4*(c*d^3 - 3*a*d*e^2)*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f(arcsin((-a/c)^ (1/4)/x), -1) + 2*(a*e^3*x + 6*a*d*e^2)*sqrt(c*x^4 + a))/(a*c*x)
Time = 1.70 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.48 \[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=e^{3} \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: c = 0 \\\frac {\sqrt {a + c x^{4}}}{2 c} & \text {otherwise} \end {cases}\right ) + \frac {3 d^{2} e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {c}} + \frac {d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {3 d e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]
e**3*Piecewise((x**4/(4*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**4)/(2*c), True )) + 3*d**2*e*asinh(sqrt(c)*x**2/sqrt(a))/(2*sqrt(c)) + d**3*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)) + 3*d*e**2*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*p i)/a)/(4*sqrt(a)*gamma(7/4))
\[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{4} + a}} \,d x } \]
\[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{\sqrt {c\,x^4+a}} \,d x \]