3.4.40 \(\int (a+\frac {b}{c+d x^2})^{3/2} \, dx\) [340]

Optimal. Leaf size=260 \[ \frac {b x \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{c}-\frac {(b-a c) x \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{c}+\frac {(b-a c) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {c} \sqrt {d} \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}}}+\frac {a \sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1985, 1986, 424, 545, 429, 506, 422} \begin {gather*} \frac {a \sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {(b-a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {c} \sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {x (b-a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{c}+\frac {b x \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2),x]

[Out]

(b*x*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c - ((b - a*c)*x*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c + ((b -
a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(Sqrt[c]*Sqrt[
d]*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]) + (a*Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*E
llipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(Sqrt[d]*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^
2))])

Rule 422

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

Rule 424

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 429

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

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 545

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 1985

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u*((b + a*c + a*d*x^n)/(c + d*x^n))^p
, x] /; FreeQ[{a, b, c, d, n, p}, x]

Rule 1986

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 10.39, size = 243, normalized size = 0.93 \begin {gather*} \frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (b \sqrt {\frac {a d}{b+a c}} x \left (b+a \left (c+d x^2\right )\right )-i a c (-b+a c) \sqrt {\frac {b+a c+a d x^2}{b+a c}} \sqrt {1+\frac {d x^2}{c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|1+\frac {b}{a c}\right )-2 i a b c \sqrt {\frac {b+a c+a d x^2}{b+a c}} \sqrt {1+\frac {d x^2}{c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|1+\frac {b}{a c}\right )\right )}{c \sqrt {\frac {a d}{b+a c}} \left (b+a \left (c+d x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 515, normalized size = 1.98

method	result	size
default	\(\frac {\left (\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {-\frac {a d}{a c +b}}\, a b d \,x^{3}+\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {a d}{a c +b}}, \sqrt {\frac {a c +b}{a c}}\right ) a^{2} c^{2}+2 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {a d}{a c +b}}, \sqrt {\frac {a c +b}{a c}}\right ) a b c -\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {a d}{a c +b}}, \sqrt {\frac {a c +b}{a c}}\right ) a b c +\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {-\frac {a d}{a c +b}}\, a b c x +\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {-\frac {a d}{a c +b}}\, b^{2} x \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {-\frac {a d}{a c +b}}\, c \left (a d \,x^{2}+a c +b \right )}\)	\(515\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(3/2),x)

[Out]

int((a + b/(c + d*x^2))^(3/2), x)

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