3.2.41 \(\int \frac {(a+b x^2)^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx\)

Optimal. Leaf size=153 \[ -\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{8 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

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Rubi [A]  time = 0.05, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1152, 416, 388, 217, 203} \begin {gather*} -\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{8 \sqrt {b} \sqrt {a^2-b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

(-9*a*x*(a - b*x^2)*Sqrt[a + b*x^2])/(8*Sqrt[a^2 - b^2*x^4]) - (x*(a - b*x^2)*(a + b*x^2)^(3/2))/(4*Sqrt[a^2 -
b^2*x^4]) + (19*a^2*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTan[(Sqrt[b]*x)/Sqrt[a - b*x^2]])/(8*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 416

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

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 )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx &=\frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {\left (a+b x^2\right )^2}{\sqrt {a-b x^2}} \, dx}{\sqrt {a^2-b^2 x^4}}\\ &=-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}-\frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {-5 a^2 b-9 a b^2 x^2}{\sqrt {a-b x^2}} \, dx}{4 b \sqrt {a^2-b^2 x^4}}\\ &=-\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {\left (19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {a-b x^2}} \, dx}{8 \sqrt {a^2-b^2 x^4}}\\ &=-\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {\left (19 a^2 \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 )}{8 \sqrt {a^2-b^2 x^4}}\\ &=-\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{8 \sqrt {b} \sqrt {a^2-b^2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 98, normalized size = 0.64 \begin {gather*} -\frac {\left (11 a x+2 b x^3\right ) \sqrt {a^2-b^2 x^4}}{8 \sqrt {a+b x^2}}+\frac {19 i a^2 \log \left (\frac {2 \sqrt {a^2-b^2 x^4}}{\sqrt {a+b x^2}}-2 i \sqrt {b} x\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

-1/8*((11*a*x + 2*b*x^3)*Sqrt[a^2 - b^2*x^4])/Sqrt[a + b*x^2] + (((19*I)/8)*a^2*Log[(-2*I)*Sqrt[b]*x + (2*Sqrt
[a^2 - b^2*x^4])/Sqrt[a + b*x^2]])/Sqrt[b]

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IntegrateAlgebraic [F]  time = 3.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2)^(5/2)/Sqrt[a^2 - b^2*x^4],x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x^2)^(5/2)/Sqrt[a^2 - b^2*x^4], x]

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fricas [A]  time = 1.13, size = 251, normalized size = 1.64 \begin {gather*} \left [-\frac {19 \, {\left (a^{2} b x^{2} + a^{3}\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 ) + 2 \, \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} + 11 \, a b x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (b^{2} x^{2} + a b\right )}}, -\frac {19 \, {\left (a^{2} b x^{2} + a^{3}\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 ) + \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} + 11 \, a b x\right )} \sqrt {b x^{2} + a}}{8 \, {\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)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*(19*(a^2*b*x^2 + a^3)*sqrt(-b)*log(-(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)) + 2*sqrt(-b^2*x^4 + a^2)*(2*b^2*x^3 + 11*a*b*x)*sqrt(b*x^2 + a))/(b^2*x^2 + a*b), -1
/8*(19*(a^2*b*x^2 + a^3)*sqrt(b)*arctan(sqrt(-b^2*x^4 + a^2)*sqrt(b*x^2 + a)*sqrt(b)/(b^2*x^3 + a*b*x)) + sqrt
(-b^2*x^4 + a^2)*(2*b^2*x^3 + 11*a*b*x)*sqrt(b*x^2 + a))/(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 {5}{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)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(5/2)/sqrt(-b^2*x^4 + a^2), x)

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maple [A]  time = 0.07, size = 132, normalized size = 0.86 \begin {gather*} -\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (2 \sqrt {-b \,x^{2}+a}\, b^{\frac {3}{2}} x^{3}-32 a^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right )+13 a^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )+11 \sqrt {-b \,x^{2}+a}\, a \sqrt {b}\, x \right )}{8 \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)^(5/2)/(-b^2*x^4+a^2)^(1/2),x)

[Out]

-1/8*(-b^2*x^4+a^2)^(1/2)*(2*x^3*b^(3/2)*(-b*x^2+a)^(1/2)+11*(-b*x^2+a)^(1/2)*b^(1/2)*x*a+13*arctan(1/(-b*x^2+
a)^(1/2)*b^(1/2)*x)*a^2-32*arctan(b^(1/2)*x/((-b*x+(a*b)^(1/2))/b*(b*x+(a*b)^(1/2)))^(1/2))*a^2)/(b*x^2+a)^(1/
2)/(-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 {5}{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)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(5/2)/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 )}^{5/2}}{\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)^(5/2)/(a^2 - b^2*x^4)^(1/2),x)

[Out]

int((a + b*x^2)^(5/2)/(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 {5}{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)**(5/2)/(-b**2*x**4+a**2)**(1/2),x)

[Out]

Integral((a + b*x**2)**(5/2)/sqrt(-(-a + b*x**2)*(a + b*x**2)), x)

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