3.285 \(\int (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\)

Optimal. Leaf size=262 \[ \frac {b \sqrt {c} e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e x (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac {e x (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6719, 413, 531, 418, 492, 411} \[ \frac {b \sqrt {c} e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e x (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac {e x (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*c - a*d)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(c*d)) + ((2*b*c - a*d)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*
x^2)])/(c*d) - ((2*b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (
b*c)/(a*d)])/(Sqrt[c]*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) + (b*Sqrt[c]*e*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))
])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 413

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

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

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

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Mathematica [C]  time = 0.32, size = 206, normalized size = 0.79 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left ((a d-b c) \left (d x \sqrt {\frac {b}{a}} \left (a+b x^2\right )-2 i b c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )+i b c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )}{c d^2 \sqrt {\frac {b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b e x^{2} + a e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{d x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 527, normalized size = 2.01 \[ \frac {\left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}} \left (d \,x^{2}+c \right ) \left (\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a b \,d^{2} x^{3}-\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, b^{2} c d \,x^{3}+\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a^{2} d^{2} x -\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a b c d x -\sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{\left (b \,x^{2}+a \right )^{2} \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, c \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)/(d*x^2+c)*e)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)

[Out]

int(((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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