3.10 \(\int \frac {(A+B x^2) (d+e x^2)}{(a+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=395 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-\sqrt {a} \sqrt {c} (A e+B d)-a B e+A c d\right )}{4 a^{5/4} c^{5/4} \sqrt {a+c x^4}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (A e+B d) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4} (A e+B d)}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {x \left (-a B e+c x^2 (A e+B d)+A c d\right )}{2 a c \sqrt {a+c x^4}}+\frac {B e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt {a+c x^4}} \]

[Out]

1/2*x*(A*c*d-a*B*e+c*(A*e+B*d)*x^2)/a/c/(c*x^4+a)^(1/2)-1/2*(A*e+B*d)*x*(c*x^4+a)^(1/2)/a/c^(1/2)/(a^(1/2)+x^2
*c^(1/2))+1/2*(A*e+B*d)*(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)/
a^(3/4)/c^(3/4)/(c*x^4+a)^(1/2)+1/2*B*e*(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))*(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^(5/4)/(c*x^4+a)^(1/2)+1/4*(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))*(a^(1/2)+x^2*c^(1/2))*(A*c*d-a*B*e-(
A*e+B*d)*a^(1/2)*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(5/4)/c^(5/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1721, 1179, 1198, 220, 1196} \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-\sqrt {a} \sqrt {c} (A e+B d)-a B e+A c d\right )}{4 a^{5/4} c^{5/4} \sqrt {a+c x^4}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (A e+B d) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4} (A e+B d)}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {x \left (-a B e+c x^2 (A e+B d)+A c d\right )}{2 a c \sqrt {a+c x^4}}+\frac {B e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(d + e*x^2))/(a + c*x^4)^(3/2),x]

[Out]

(x*(A*c*d - a*B*e + c*(B*d + A*e)*x^2))/(2*a*c*Sqrt[a + c*x^4]) - ((B*d + A*e)*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]
*(Sqrt[a] + Sqrt[c]*x^2)) + ((B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E
llipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*c^(3/4)*Sqrt[a + c*x^4]) + (B*e*(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^(5/
4)*Sqrt[a + c*x^4]) + ((A*c*d - a*B*e - Sqrt[a]*Sqrt[c]*(B*d + A*e))*(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])/(4*a^(5/4)*c^(5/4)*Sqrt[a + c*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 1721

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a +
c*x^4], Px*(d + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x^2] && NeQ[c*d^
2 + a*e^2, 0] && IntegerQ[p + 1/2] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )}{\left (a+c x^4\right )^{3/2}} \, dx &=\int \left (\frac {A c d-a B e+c (B d+A e) x^2}{c \left (a+c x^4\right )^{3/2}}+\frac {B e}{c \sqrt {a+c x^4}}\right ) \, dx\\ &=\frac {\int \frac {A c d-a B e+c (B d+A e) x^2}{\left (a+c x^4\right )^{3/2}} \, dx}{c}+\frac {(B e) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{c}\\ &=\frac {x \left (A c d-a B e+c (B d+A e) x^2\right )}{2 a c \sqrt {a+c x^4}}+\frac {B e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt {a+c x^4}}-\frac {\int \frac {-A c d+a B e+c (B d+A e) x^2}{\sqrt {a+c x^4}} \, dx}{2 a c}\\ &=\frac {x \left (A c d-a B e+c (B d+A e) x^2\right )}{2 a c \sqrt {a+c x^4}}+\frac {B e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt {a+c x^4}}+\frac {(B d+A e) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{2 \sqrt {a} \sqrt {c}}+\frac {\left (A c d-a B e-\sqrt {a} \sqrt {c} (B d+A e)\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{2 a c}\\ &=\frac {x \left (A c d-a B e+c (B d+A e) x^2\right )}{2 a c \sqrt {a+c x^4}}-\frac {(B d+A e) x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {(B d+A e) \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 )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}+\frac {B e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt {a+c x^4}}+\frac {\left (A c d-a B e-\sqrt {a} \sqrt {c} (B d+A e)\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} c^{5/4} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 126, normalized size = 0.32 \[ \frac {3 x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right ) (a B e+A c d)+2 c x^3 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {c x^4}{a}\right ) (A e+B d)+3 x (A c d-a B e)}{6 a c \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(d + e*x^2))/(a + c*x^4)^(3/2),x]

[Out]

(3*(A*c*d - a*B*e)*x + 3*(A*c*d + a*B*e)*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)]
+ 2*c*(B*d + A*e)*x^3*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^4)/a)])/(6*a*c*Sqrt[a + c*x^
4])

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fricas [F]  time = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{4} + {\left (B d + A e\right )} x^{2} + A d\right )} \sqrt {c x^{4} + a}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral((B*e*x^4 + (B*d + A*e)*x^2 + A*d)*sqrt(c*x^4 + a)/(c^2*x^8 + 2*a*c*x^4 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)/(c*x^4 + a)^(3/2), x)

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maple [C]  time = 0.00, size = 320, normalized size = 0.81 \[ \left (\frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}\right ) A d +\left (-\frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, c}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c}\right ) B e +\left (A e +B d \right ) \left (\frac {x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}\, \sqrt {c}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(e*x^2+d)/(c*x^4+a)^(3/2),x)

[Out]

B*e*(-1/2/((x^4+a/c)*c)^(1/2)/c*x+1/2/c/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*
c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I))+(A*e+B*d)*(1/2/((x^4+a/c)*c)^(1
/2)/a*x^3-1/2*I/a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(
1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I
)))+A*d*(1/2/((x^4+a/c)*c)^(1/2)/a*x+1/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/
2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)/(c*x^4 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,\left (e\,x^2+d\right )}{{\left (c\,x^4+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(d + e*x^2))/(a + c*x^4)^(3/2),x)

[Out]

int(((A + B*x^2)*(d + e*x^2))/(a + c*x^4)^(3/2), x)

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sympy [C]  time = 12.95, size = 167, normalized size = 0.42 \[ \frac {A d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {A e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B e x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(e*x**2+d)/(c*x**4+a)**(3/2),x)

[Out]

A*d*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + A*e*x**3*gamma(
3/4)*hyper((3/4, 3/2), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7/4)) + B*d*x**3*gamma(3/4)*hyper((
3/4, 3/2), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7/4)) + B*e*x**5*gamma(5/4)*hyper((5/4, 3/2), (
9/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(9/4))

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