3.842 \(\int \frac {(a+b x^2)^2}{\sqrt {e x} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=193 \[ \frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (21 a^2 d^2-14 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{c} d^{9/4} \sqrt {e} \sqrt {c+d x^2}}-\frac {2 b \sqrt {e x} \sqrt {c+d x^2} (5 b c-14 a d)}{21 d^2 e}+\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3} \]

[Out]

2/7*b^2*(e*x)^(5/2)*(d*x^2+c)^(1/2)/d/e^3-2/21*b*(-14*a*d+5*b*c)*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d^2/e+1/21*(21*a^
2*d^2-14*a*b*c*d+5*b^2*c^2)*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*
(e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/
2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^(1/4)/d^(9/4)/e^(1/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {464, 459, 329, 220} \[ \frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (21 a^2 d^2-14 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{c} d^{9/4} \sqrt {e} \sqrt {c+d x^2}}-\frac {2 b \sqrt {e x} \sqrt {c+d x^2} (5 b c-14 a d)}{21 d^2 e}+\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(5*b*c - 14*a*d)*Sqrt[e*x]*Sqrt[c + d*x^2])/(21*d^2*e) + (2*b^2*(e*x)^(5/2)*Sqrt[c + d*x^2])/(7*d*e^3) +
((5*b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*Ellipt
icF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(21*c^(1/4)*d^(9/4)*Sqrt[e]*Sqrt[c + d*x^2])

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x^2}} \, dx &=\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3}+\frac {2 \int \frac {\frac {7 a^2 d}{2}-\frac {1}{2} b (5 b c-14 a d) x^2}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{7 d}\\ &=-\frac {2 b (5 b c-14 a d) \sqrt {e x} \sqrt {c+d x^2}}{21 d^2 e}+\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3}-\frac {1}{21} \left (-21 a^2-\frac {b c (5 b c-14 a d)}{d^2}\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx\\ &=-\frac {2 b (5 b c-14 a d) \sqrt {e x} \sqrt {c+d x^2}}{21 d^2 e}+\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3}+\frac {\left (2 \left (21 a^2+\frac {b c (5 b c-14 a d)}{d^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{21 e}\\ &=-\frac {2 b (5 b c-14 a d) \sqrt {e x} \sqrt {c+d x^2}}{21 d^2 e}+\frac {2 b^2 (e x)^{5/2} \sqrt {c+d x^2}}{7 d e^3}+\frac {\left (21 a^2+\frac {b c (5 b c-14 a d)}{d^2}\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{c} \sqrt [4]{d} \sqrt {e} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 148, normalized size = 0.77 \[ \frac {2 x \left (-b \left (c+d x^2\right ) \left (-14 a d+5 b c-3 b d x^2\right )+\frac {i \sqrt {x} \sqrt {\frac {c}{d x^2}+1} \left (21 a^2 d^2-14 a b c d+5 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}\right )}{21 d^2 \sqrt {e x} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-(b*(c + d*x^2)*(5*b*c - 14*a*d - 3*b*d*x^2)) + (I*(5*b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[1 + c/(d*x
^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/Sqrt[(I*Sqrt[c])/Sqrt[d]]))/(21*d^2*
Sqrt[e*x]*Sqrt[c + d*x^2])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d e x^{3} + c e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(d*e*x^3 + c*e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 350, normalized size = 1.81 \[ \frac {6 b^{2} d^{3} x^{5}+28 a b \,d^{3} x^{3}-4 b^{2} c \,d^{2} x^{3}+28 a b c \,d^{2} x -10 b^{2} c^{2} d x +21 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-14 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b c d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{21 \sqrt {d \,x^{2}+c}\, \sqrt {e x}\, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(e*x)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

1/21/(d*x^2+c)^(1/2)*(21*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*((d*x+(-c*d)^(1/2))/(-
c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*(-c*d)^(1/2)*a^
2*d^2-14*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*(-c*d)^(1/2)*a*b*c*d+5*Elliptic
F(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+
(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*(-c*d)^(1/2)*b^2*c^2+6*x^5*b^2*d^3+28*x^3*a*b*d^
3-4*x^3*b^2*c*d^2+28*x*a*b*c*d^2-10*x*b^2*c^2*d)/(e*x)^(1/2)/d^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^2}{\sqrt {e\,x}\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/((e*x)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(1/2)*(c + d*x^2)^(1/2)), x)

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sympy [C]  time = 5.31, size = 144, normalized size = 0.75 \[ \frac {a^{2} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {a b x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} + \frac {b^{2} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \sqrt {e} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(e*x)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

a**2*sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*sqrt(e)*gamma(5/4)) + a
*b*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*sqrt(e)*gamma(9/4)) + b**2
*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*sqrt(e)*gamma(13/4))

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