3.2.42 \(\int \frac {(a+b x^2)^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx\)

Optimal. Leaf size=110 \[ \frac {3 a \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{2 \sqrt {b} \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \sqrt {a+b x^2}}{2 \sqrt {a^2-b^2 x^4}} \]

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Rubi [A]  time = 0.03, antiderivative size = 110, 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 = {1152, 388, 217, 203} \begin {gather*} \frac {3 a \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{2 \sqrt {b} \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \sqrt {a+b x^2}}{2 \sqrt {a^2-b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

-(x*(a - b*x^2)*Sqrt[a + b*x^2])/(2*Sqrt[a^2 - b^2*x^4]) + (3*a*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTan[(Sqrt[b
]*x)/Sqrt[a - b*x^2]])/(2*Sqrt[b]*Sqrt[a^2 - b^2*x^4])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 388

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

Rule 1152

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^FracPart[p]/((d + e*x
^2)^FracPart[p]*(a/d + (c*x^2)/e)^FracPart[p]), Int[(d + e*x^2)^(p + q)*(a/d + (c*x^2)/e)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 86, normalized size = 0.78 \begin {gather*} -\frac {x \sqrt {a^2-b^2 x^4}}{2 \sqrt {a+b x^2}}+\frac {3 i a \log \left (\frac {2 \sqrt {a^2-b^2 x^4}}{\sqrt {a+b x^2}}-2 i \sqrt {b} x\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

-1/2*(x*Sqrt[a^2 - b^2*x^4])/Sqrt[a + b*x^2] + (((3*I)/2)*a*Log[(-2*I)*Sqrt[b]*x + (2*Sqrt[a^2 - b^2*x^4])/Sqr
t[a + b*x^2]])/Sqrt[b]

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IntegrateAlgebraic [F]  time = 2.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right )^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2)^(3/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x^2)^(3/2)/Sqrt[a^2 - b^2*x^4], x]

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fricas [A]  time = 1.09, size = 223, normalized size = 2.03 \begin {gather*} \left [-\frac {2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} b x + 3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {-b} \log \left (-\frac {2 \, b^{2} x^{4} + a b x^{2} - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b x^{2} + a}\right )}{4 \, {\left (b^{2} x^{2} + a b\right )}}, -\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} b x + 3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{b^{2} x^{3} + a b x}\right )}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(2*sqrt(-b^2*x^4 + a^2)*sqrt(b*x^2 + a)*b*x + 3*(a*b*x^2 + a^2)*sqrt(-b)*log(-(2*b^2*x^4 + a*b*x^2 - 2*s
qrt(-b^2*x^4 + a^2)*sqrt(b*x^2 + a)*sqrt(-b)*x - a^2)/(b*x^2 + a)))/(b^2*x^2 + a*b), -1/2*(sqrt(-b^2*x^4 + a^2
)*sqrt(b*x^2 + a)*b*x + 3*(a*b*x^2 + a^2)*sqrt(b)*arctan(sqrt(-b^2*x^4 + a^2)*sqrt(b*x^2 + a)*sqrt(b)/(b^2*x^3
+ a*b*x)))/(b^2*x^2 + a*b)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\sqrt {-b^{2} x^{4} + a^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)/sqrt(-b^2*x^4 + a^2), x)

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maple [A]  time = 0.02, size = 107, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (-4 a \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right )+a \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )+\sqrt {-b \,x^{2}+a}\, \sqrt {b}\, x \right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x)

[Out]

-1/2/(b*x^2+a)^(1/2)*(-b^2*x^4+a^2)^(1/2)*(x*(-b*x^2+a)^(1/2)*b^(1/2)+a*arctan(1/(-b*x^2+a)^(1/2)*b^(1/2)*x)-4
*arctan(1/((-b*x+(a*b)^(1/2))*(b*x+(a*b)^(1/2))/b)^(1/2)*b^(1/2)*x)*a)/(-b*x^2+a)^(1/2)/b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\sqrt {-b^{2} x^{4} + a^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/sqrt(-b^2*x^4 + a^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{\sqrt {a^2-b^2\,x^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(a^2 - b^2*x^4)^(1/2),x)

[Out]

int((a + b*x^2)^(3/2)/(a^2 - b^2*x^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(-b**2*x**4+a**2)**(1/2),x)

[Out]

Integral((a + b*x**2)**(3/2)/sqrt(-(-a + b*x**2)*(a + b*x**2)), x)

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