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

Optimal. Leaf size=267 \[ \frac {c^{3/2} \sqrt {d} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a b \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {d x \sqrt {a+b x^2} (b c-2 a d)}{a b^2 \sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} (b c-2 a d) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b^2 \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {c+d x^2} (b c-a d)}{a b \sqrt {a+b x^2}} \]

[Out]

-d*(-2*a*d+b*c)*x*(b*x^2+a)^(1/2)/a/b^2/(d*x^2+c)^(1/2)+c^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*Ellipt
icF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*d^(1/2)*(b*x^2+a)^(1/2)/a/b/(c*(b*x^2+a)/a/(d*x^2+c
))^(1/2)/(d*x^2+c)^(1/2)+(-2*a*d+b*c)*(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))*c^(1/2)*d^(1/2)*(b*x^2+a)^(1/2)/a/b^2/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+
c)^(1/2)+(-a*d+b*c)*x*(d*x^2+c)^(1/2)/a/b/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {413, 531, 418, 492, 411} \[ -\frac {d x \sqrt {a+b x^2} (b c-2 a d)}{a b^2 \sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} (b c-2 a d) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b^2 \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {c^{3/2} \sqrt {d} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {c+d x^2} (b c-a d)}{a b \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.30, size = 191, normalized size = 0.72 \[ \frac {(b c-a d) \left (x \sqrt {\frac {b}{a}} \left (c+d x^2\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \operatorname {EllipticF}\left (i \sinh ^{-1}\left (x \sqrt {\frac {b}{a}}\right ),\frac {a d}{b c}\right )\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (2 a d-b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{a^2 \left (\frac {b}{a}\right )^{3/2} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 332, normalized size = 1.24 \[ \frac {\left (-\sqrt {-\frac {b}{a}}\, a \,d^{2} x^{3}+\sqrt {-\frac {b}{a}}\, b c d \,x^{3}-\sqrt {-\frac {b}{a}}\, a c d x +2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, b \,c^{2} x -\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right ) \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{\left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right ) \sqrt {-\frac {b}{a}}\, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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