Integrand size = 19, antiderivative size = 355 \[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}+\frac {3 a d^2 e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}-\frac {6 a^{5/4} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{3/4} d \left (5 \sqrt {c} d^2+9 \sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}} \]
1/6*e^3*(c*x^4+a)^(3/2)/c+3/4*a*d^2*e*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2)) /c^(1/2)+3/4*d^2*e*x^2*(c*x^4+a)^(1/2)+1/15*d*x*(9*e^2*x^2+5*d^2)*(c*x^4+a )^(1/2)+6/5*a*d*e^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-6/5*a^ (5/4)*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)+1/15*a^(3/4)*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))*(9*e^2*a^(1/2)+5*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)/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 = 186, normalized size of antiderivative = 0.52 \[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\frac {\sqrt {a+c x^4} \left (2 a e^3 \sqrt {1+\frac {c x^4}{a}}+9 c d^2 e x^2 \sqrt {1+\frac {c x^4}{a}}+2 c e^3 x^4 \sqrt {1+\frac {c x^4}{a}}+9 \sqrt {a} \sqrt {c} d^2 e \text {arcsinh}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )+12 c d^3 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^4}{a}\right )+12 c d e^2 x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )\right )}{12 c \sqrt {1+\frac {c x^4}{a}}} \]
(Sqrt[a + c*x^4]*(2*a*e^3*Sqrt[1 + (c*x^4)/a] + 9*c*d^2*e*x^2*Sqrt[1 + (c* x^4)/a] + 2*c*e^3*x^4*Sqrt[1 + (c*x^4)/a] + 9*Sqrt[a]*Sqrt[c]*d^2*e*ArcSin h[(Sqrt[c]*x^2)/Sqrt[a]] + 12*c*d^3*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -( (c*x^4)/a)] + 12*c*d*e^2*x^3*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^4)/a )]))/(12*c*Sqrt[1 + (c*x^4)/a])
Time = 0.47 (sec) , antiderivative size = 355, 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.
| \(\displaystyle \int \sqrt {a+c x^4} (d+e x)^3 \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\sqrt {a+c x^4} \left (d^3+3 d e^2 x^2\right )+x \sqrt {a+c x^4} \left (3 d^2 e+e^3 x^2\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{3/4} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (9 \sqrt {a} e^2+5 \sqrt {c} d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {6 a^{5/4} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {3 a d^2 e \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}+\frac {1}{15} d x \sqrt {a+c x^4} \left (5 d^2+9 e^2 x^2\right )+\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}\) |
(3*d^2*e*x^2*Sqrt[a + c*x^4])/4 + (6*a*d*e^2*x*Sqrt[a + c*x^4])/(5*Sqrt[c] *(Sqrt[a] + Sqrt[c]*x^2)) + (d*x*(5*d^2 + 9*e^2*x^2)*Sqrt[a + c*x^4])/15 + (e^3*(a + c*x^4)^(3/2))/(6*c) + (3*a*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(4*Sqrt[c]) - (6*a^(5/4)*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])/(5*c^(3/4)*Sqrt[a + c*x^4]) + (a^(3/4)*d*(5*Sqrt[c]*d^2 + 9*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 ])
3.3.6.3.1 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 = 1.73 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.74
| method | result | size |
| risch | \(\frac {\left (10 e^{3} x^{4} c +36 d \,e^{2} x^{3} c +45 c \,d^{2} e \,x^{2}+20 d^{3} c x +10 a \,e^{3}\right ) \sqrt {c \,x^{4}+a}}{60 c}+\frac {d a \left (\frac {20 d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {36 i e^{2} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {45 e d \ln \left (x^{2} \sqrt {c}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}\right )}{30}\) | \(261\) |
| default | \(d^{3} \left (\frac {x \sqrt {c \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+\frac {e^{3} \left (c \,x^{4}+a \right )^{\frac {3}{2}}}{6 c}+3 d \,e^{2} \left (\frac {x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\right )+3 d^{2} e \left (\frac {x^{2} \sqrt {c \,x^{4}+a}}{4}+\frac {a \ln \left (x^{2} \sqrt {c}+\sqrt {c \,x^{4}+a}\right )}{4 \sqrt {c}}\right )\) | \(269\) |
| elliptic | \(\frac {e^{3} x^{4} \sqrt {c \,x^{4}+a}}{6}+\frac {3 d \,e^{2} x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {3 d^{2} e \,x^{2} \sqrt {c \,x^{4}+a}}{4}+\frac {d^{3} x \sqrt {c \,x^{4}+a}}{3}+\frac {e^{3} a \sqrt {c \,x^{4}+a}}{6 c}+\frac {2 a \,d^{3} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {3 a \,d^{2} e \ln \left (2 x^{2} \sqrt {c}+2 \sqrt {c \,x^{4}+a}\right )}{4 \sqrt {c}}+\frac {6 i d \,e^{2} a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(293\) |
1/60*(10*c*e^3*x^4+36*c*d*e^2*x^3+45*c*d^2*e*x^2+20*c*d^3*x+10*a*e^3)/c*(c *x^4+a)^(1/2)+1/30*d*a*(20*d^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)*EllipticF(x *(I/a^(1/2)*c^(1/2))^(1/2),I)+36*I*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))+45/2*e*d*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))/c^(1/2))
Time = 0.16 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.53 \[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\frac {144 \, a \sqrt {c} d e^{2} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 45 \, a \sqrt {c} d^{2} e x \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 16 \, {\left (5 \, c d^{3} - 9 \, a d e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (10 \, c e^{3} x^{5} + 36 \, c d e^{2} x^{4} + 45 \, c d^{2} e x^{3} + 20 \, c d^{3} x^{2} + 10 \, a e^{3} x + 72 \, a d e^{2}\right )} \sqrt {c x^{4} + a}}{120 \, c x} \]
1/120*(144*a*sqrt(c)*d*e^2*x*(-a/c)^(3/4)*elliptic_e(arcsin((-a/c)^(1/4)/x ), -1) + 45*a*sqrt(c)*d^2*e*x*log(-2*c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 16*(5*c*d^3 - 9*a*d*e^2)*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f(arcsin( (-a/c)^(1/4)/x), -1) + 2*(10*c*e^3*x^5 + 36*c*d*e^2*x^4 + 45*c*d^2*e*x^3 + 20*c*d^3*x^2 + 10*a*e^3*x + 72*a*d*e^2)*sqrt(c*x^4 + a))/(c*x)
Time = 2.01 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.49 \[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\frac {\sqrt {a} d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 \sqrt {a} d^{2} e x^{2} \sqrt {1 + \frac {c x^{4}}{a}}}{4} + \frac {3 \sqrt {a} d e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 a d^{2} e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {c}} + e^{3} \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: c = 0 \\\frac {\left (a + c x^{4}\right )^{\frac {3}{2}}}{6 c} & \text {otherwise} \end {cases}\right ) \]
sqrt(a)*d**3*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**4*exp_polar(I*pi )/a)/(4*gamma(5/4)) + 3*sqrt(a)*d**2*e*x**2*sqrt(1 + c*x**4/a)/4 + 3*sqrt( a)*d*e**2*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi )/a)/(4*gamma(7/4)) + 3*a*d**2*e*asinh(sqrt(c)*x**2/sqrt(a))/(4*sqrt(c)) + e**3*Piecewise((sqrt(a)*x**4/4, Eq(c, 0)), ((a + c*x**4)**(3/2)/(6*c), Tr ue))
\[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\int { \sqrt {c x^{4} + a} {\left (e x + d\right )}^{3} \,d x } \]
\[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\int { \sqrt {c x^{4} + a} {\left (e x + d\right )}^{3} \,d x } \]
Timed out. \[ \int (d+e x)^3 \sqrt {a+c x^4} \, dx=\int \sqrt {c\,x^4+a}\,{\left (d+e\,x\right )}^3 \,d x \]