3.184 \(\int (d+e x)^2 \sqrt{a+c x^4} \, dx\)

Optimal. Leaf size=326 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+5 \sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{a+c x^4}}-\frac{2 a^{5/4} e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{1}{15} x \sqrt{a+c x^4} \left (5 d^2+3 e^2 x^2\right )+\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{a d e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

[Out]

(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[(Sqr
t[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])

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Rubi [A]  time = 0.403576, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (3 \sqrt{a} e^2+5 \sqrt{c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{a+c x^4}}-\frac{2 a^{5/4} e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{a+c x^4}}+\frac{1}{15} x \sqrt{a+c x^4} \left (5 d^2+3 e^2 x^2\right )+\frac{1}{2} d e x^2 \sqrt{a+c x^4}+\frac{a d e \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{2 \sqrt{c}}+\frac{2 a e^2 x \sqrt{a+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

Antiderivative was successfully verified.

[In]  Int[(d + e*x)^2*Sqrt[a + c*x^4],x]

[Out]

(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[(Sqr
t[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])

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Rubi in Sympy [A]  time = 44.4783, size = 301, normalized size = 0.92 \[ - \frac{2 a^{\frac{5}{4}} e^{2} \sqrt{\frac{a + c x^{4}}{\left (\sqrt{a} + \sqrt{c} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x^{2}\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 c^{\frac{3}{4}} \sqrt{a + c x^{4}}} + \frac{a^{\frac{3}{4}} \sqrt{\frac{a + c x^{4}}{\left (\sqrt{a} + \sqrt{c} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x^{2}\right ) \left (3 \sqrt{a} e^{2} + 5 \sqrt{c} d^{2}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 c^{\frac{3}{4}} \sqrt{a + c x^{4}}} + \frac{a d e \operatorname{atanh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a + c x^{4}}} \right )}}{2 \sqrt{c}} + \frac{2 a e^{2} x \sqrt{a + c x^{4}}}{5 \sqrt{c} \left (\sqrt{a} + \sqrt{c} x^{2}\right )} + \frac{d e x^{2} \sqrt{a + c x^{4}}}{2} + \frac{x \sqrt{a + c x^{4}} \left (5 d^{2} + 3 e^{2} x^{2}\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x+d)**2*(c*x**4+a)**(1/2),x)

[Out]

-2*a**(5/4)*e**2*sqrt((a + c*x**4)/(sqrt(a) + sqrt(c)*x**2)**2)*(sqrt(a) + sqrt(
c)*x**2)*elliptic_e(2*atan(c**(1/4)*x/a**(1/4)), 1/2)/(5*c**(3/4)*sqrt(a + c*x**
4)) + a**(3/4)*sqrt((a + c*x**4)/(sqrt(a) + sqrt(c)*x**2)**2)*(sqrt(a) + sqrt(c)
*x**2)*(3*sqrt(a)*e**2 + 5*sqrt(c)*d**2)*elliptic_f(2*atan(c**(1/4)*x/a**(1/4)),
1/2)/(15*c**(3/4)*sqrt(a + c*x**4)) + a*d*e*atanh(sqrt(c)*x**2/sqrt(a + c*x**4)
)/(2*sqrt(c)) + 2*a*e**2*x*sqrt(a + c*x**4)/(5*sqrt(c)*(sqrt(a) + sqrt(c)*x**2))
+ d*e*x**2*sqrt(a + c*x**4)/2 + x*sqrt(a + c*x**4)*(5*d**2 + 3*e**2*x**2)/15

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Mathematica [C]  time = 0.699326, size = 247, normalized size = 0.76 \[ \frac{12 a^{3/2} e^2 \sqrt{\frac{c x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )-4 a \sqrt{\frac{c x^4}{a}+1} \left (3 \sqrt{a} e^2+5 i \sqrt{c} d^2\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )+\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} \left (\sqrt{c} x \left (a+c x^4\right ) \left (10 d^2+15 d e x+6 e^2 x^2\right )+15 a d e \sqrt{a+c x^4} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )\right )}{30 \sqrt{c} \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[(d + e*x)^2*Sqrt[a + c*x^4],x]

[Out]

(Sqrt[(I*Sqrt[c])/Sqrt[a]]*(Sqrt[c]*x*(10*d^2 + 15*d*e*x + 6*e^2*x^2)*(a + c*x^4
) + 15*a*d*e*Sqrt[a + c*x^4]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]]) + 12*a^(3/2
)*e^2*Sqrt[1 + (c*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1]
- 4*a*((5*I)*Sqrt[c]*d^2 + 3*Sqrt[a]*e^2)*Sqrt[1 + (c*x^4)/a]*EllipticF[I*ArcSin
h[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/(30*Sqrt[(I*Sqrt[c])/Sqrt[a]]*Sqrt[c]*Sqrt[
a + c*x^4])

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Maple [C]  time = 0.01, size = 310, normalized size = 1. \[{\frac{{d}^{2}x}{3}\sqrt{c{x}^{4}+a}}+{\frac{2\,a{d}^{2}}{3}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}}+{\frac{{e}^{2}{x}^{3}}{5}\sqrt{c{x}^{4}+a}}+{{\frac{2\,i}{5}}{e}^{2}{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}}-{{\frac{2\,i}{5}}{e}^{2}{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}}+{\frac{de{x}^{2}}{2}\sqrt{c{x}^{4}+a}}+{\frac{ade}{2}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){\frac{1}{\sqrt{c}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x+d)^2*(c*x^4+a)^(1/2),x)

[Out]

1/3*d^2*x*(c*x^4+a)^(1/2)+2/3*d^2*a/(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)+1/5*e^2*x^3*(c*x^4+a)^(1/2)+2/5*I*e^2*a^(3/2)/(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)-2/5*I*e^2*a^(3/2
)/(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)*EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+1/2
*d*e*x^2*(c*x^4+a)^(1/2)+1/2*d*e*a/c^(1/2)*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \sqrt{c x^{4} + a}{\left (e x + d\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^4 + a)*(e*x + d)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + a)*(e*x + d)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\sqrt{c x^{4} + a}{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^4 + a)*(e*x + d)^2,x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + a)*(e^2*x^2 + 2*d*e*x + d^2), x)

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Sympy [A]  time = 9.74102, size = 138, normalized size = 0.42 \[ \frac{\sqrt{a} d^{2} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} d e x^{2} \sqrt{1 + \frac{c x^{4}}{a}}}{2} + \frac{\sqrt{a} e^{2} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{a d e \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x+d)**2*(c*x**4+a)**(1/2),x)

[Out]

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*gamma(7/4)) + a*d*e
*asinh(sqrt(c)*x**2/sqrt(a))/(2*sqrt(c))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \sqrt{c x^{4} + a}{\left (e x + d\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^4 + a)*(e*x + d)^2,x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + a)*(e*x + d)^2, x)