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

Optimal. Leaf size=213 \[ \frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 c^{9/4} d^{9/4} \sqrt {e} \sqrt {c+d x^2}}-\frac {\sqrt {e x} (5 a d+7 b c) (b c-a d)}{6 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\sqrt {e x} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}} \]

[Out]

1/3*(-a*d+b*c)^2*(e*x)^(1/2)/c/d^2/e/(d*x^2+c)^(3/2)-1/6*(-a*d+b*c)*(5*a*d+7*b*c)*(e*x)^(1/2)/c^2/d^2/e/(d*x^2
+c)^(1/2)+1/12*(5*a^2*d^2+2*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)/co
s(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^(9/4)/d^(9/4)/e^(1/2)/(d*x^2+c)^(1/
2)

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Rubi [A]  time = 0.16, antiderivative size = 213, 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 = {463, 457, 329, 220} \[ \frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 c^{9/4} d^{9/4} \sqrt {e} \sqrt {c+d x^2}}-\frac {\sqrt {e x} (5 a d+7 b c) (b c-a d)}{6 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\sqrt {e x} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[e*x])/(3*c*d^2*e*(c + d*x^2)^(3/2)) - ((b*c - a*d)*(7*b*c + 5*a*d)*Sqrt[e*x])/(6*c^2*d^2*e
*Sqrt[c + d*x^2]) + ((5*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqr
t[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(9/4)*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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 463

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \left (c+d x^2\right )^{5/2}} \, dx &=\frac {(b c-a d)^2 \sqrt {e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {\frac {1}{2} \left (-6 a^2 d^2+(b c-a d)^2\right )-3 b^2 c d x^2}{\sqrt {e x} \left (c+d x^2\right )^{3/2}} \, dx}{3 c d^2}\\ &=\frac {(b c-a d)^2 \sqrt {e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (7 b c+5 a d) \sqrt {e x}}{6 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{12 c^2 d^2}\\ &=\frac {(b c-a d)^2 \sqrt {e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (7 b c+5 a d) \sqrt {e x}}{6 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{6 c^2 d^2 e}\\ &=\frac {(b c-a d)^2 \sqrt {e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (7 b c+5 a d) \sqrt {e x}}{6 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2+2 a b c d+5 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 )}{12 c^{9/4} d^{9/4} \sqrt {e} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 169, normalized size = 0.79 \[ \frac {x \left (\frac {i \sqrt {x} \sqrt {\frac {c}{d x^2}+1} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}+5 a^2 d^2+\frac {2 c (b c-a d)^2}{c+d x^2}+2 a b c d-7 b^2 c^2\right )}{6 c^2 d^2 \sqrt {e x} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-7*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2 + (2*c*(b*c - a*d)^2)/(c + d*x^2) + (I*(5*b^2*c^2 + 2*a*b*c*d + 5*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]]))/(6*c^2*d^2*Sqrt[e*x]*Sqrt[c + d*x^2])

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d^{3} e x^{7} + 3 \, c d^{2} e x^{5} + 3 \, c^{2} d e x^{3} + c^{3} e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(5/2)/(e*x)^(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^3*e*x^7 + 3*c*d^2*e*x^5 + 3*c^2*d*e*x^3 + c^
3*e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 660, normalized size = 3.10 \[ \frac {10 a^{2} d^{4} x^{3}+4 a b c \,d^{3} x^{3}-14 b^{2} c^{2} d^{2} x^{3}+5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} d^{3} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b c \,d^{2} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{2} d \,x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+14 a^{2} c \,d^{3} x -4 a b \,c^{2} d^{2} x -10 b^{2} c^{3} d x +5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b \,c^{2} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{3} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{12 \sqrt {e x}\, \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(5*((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))*(-c*d)^(1/2)*x^2*a^2*d^3+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))*(-c*d)^(1/2)*x^2*a*b*c*d^2+5*((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))*(-c*d)^(1/2)*x^2*b^2*c^2*d+5*((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))*(-c*d)^(1/2)*a^2*c*d^2+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))*(-c*d)^(1/2)*a*b*c^2*d+5*((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))*(-c*d)^(1/2)*b^2*c^3+10*a^2*d^4*x^3+4*a*b*c*d^3*x^3-14*b^2*c^2*d^2*x^3+14*a^2*c*d^3*x-4*a*b*c^2*d
^2*x-10*b^2*c^3*d*x)/(e*x)^(1/2)/c^2/d^3/(d*x^2+c)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)^(5/2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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