3.844 \(\int \frac {(a+b x^2)^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a^2 d^2-6 a b c d+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3 c^{5/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}}-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3} \]

[Out]

-2/3*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(3/2)+2/3*b^2*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d/e^3-1/3*(a^2*d^2-6*a*b*c*d+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^(5/4)/d^(5/4)/e^(5/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 184, 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 = {462, 459, 329, 220} \[ -\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a^2 d^2-6 a b c d+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3 c^{5/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}}-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(3*c*e*(e*x)^(3/2)) + (2*b^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(3*d*e^3) - ((b^2*c^2 - 6*a*b
*c*d + a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sq
rt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(3*c^(5/4)*d^(5/4)*e^(5/2)*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 462

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x^2}} \, dx &=-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 \int \frac {\frac {1}{2} a (6 b c-a d)+\frac {3}{2} b^2 c x^2}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{3 c e^2}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3}-\frac {\left (b^2 c^2-6 a b c d+a^2 d^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{3 c d e^2}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3}-\frac {\left (2 \left (b^2 c^2-6 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 c d e^3}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \sqrt {c+d x^2}}{3 d e^3}-\frac {\left (b^2 c^2-6 a b c d+a^2 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 )}{3 c^{5/4} d^{5/4} e^{5/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 165, normalized size = 0.90 \[ \frac {x \left (2 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} \left (c+d x^2\right ) \left (b^2 c x^2-a^2 d\right )-2 i x^{5/2} \sqrt {\frac {c}{d x^2}+1} \left (a^2 d^2-6 a b c d+b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )\right )}{3 c d \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} (e x)^{5/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

(x*(2*Sqrt[(I*Sqrt[c])/Sqrt[d]]*(-(a^2*d) + b^2*c*x^2)*(c + d*x^2) - (2*I)*(b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Sqr
t[1 + c/(d*x^2)]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1]))/(3*c*Sqrt[(I*Sqrt[c])/S
qrt[d]]*d*(e*x)^(5/2)*Sqrt[c + d*x^2])

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d e^{3} x^{5} + c e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(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^3*x^5 + c*e^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(5/2)), x)

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maple [A]  time = 0.02, size = 352, normalized size = 1.91 \[ -\frac {-2 b^{2} c \,d^{2} x^{4}+2 a^{2} d^{3} x^{2}-2 b^{2} c^{2} d \,x^{2}+\sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} d^{2} x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b c d x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{2} x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+2 a^{2} c \,d^{2}}{3 \sqrt {d \,x^{2}+c}\, \sqrt {e x}\, c \,d^{2} e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/3/(d*x^2+c)^(1/2)/x*((-c*d)^(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)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*a
^2*d^2-6*(-c*d)^(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)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*a*b*c*d+(-c*d)^(
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)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*b^2*c^2-2*b^2*c*d^2*x^4+2*a^2*d^3
*x^2-2*b^2*c^2*d*x^2+2*a^2*c*d^2)/c/e^2/(e*x)^(1/2)/d^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{5/2}\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(1/2)), x)

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sympy [C]  time = 12.60, size = 148, normalized size = 0.80 \[ \frac {a^{2} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {a b \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{2} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

a**2*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e**(5/2)*x**(3/2)*gamma(1/4))
+ a*b*sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*e**(5/2)*gamma(5/4)) +
b**2*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e**(5/2)*gamma(9/4))

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