3.1.66 \(\int \frac {1}{(a+b \cos ^2(x))^{3/2}} \, dx\) [66]

Optimal. Leaf size=78 \[ \frac {\sqrt {a+b \cos ^2(x)} E\left (\frac {\pi }{2}+x|-\frac {b}{a}\right )}{a (a+b) \sqrt {1+\frac {b \cos ^2(x)}{a}}}-\frac {b \cos (x) \sin (x)}{a (a+b) \sqrt {a+b \cos ^2(x)}} \]

[Out]

-b*cos(x)*sin(x)/a/(a+b)/(a+b*cos(x)^2)^(1/2)-(sin(x)^2)^(1/2)/sin(x)*EllipticE(cos(x),(-b/a)^(1/2))*(a+b*cos(
x)^2)^(1/2)/a/(a+b)/(1+b*cos(x)^2/a)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3263, 21, 3257, 3256} \begin {gather*} \frac {\sqrt {a+b \cos ^2(x)} E\left (x+\frac {\pi }{2}|-\frac {b}{a}\right )}{a (a+b) \sqrt {\frac {b \cos ^2(x)}{a}+1}}-\frac {b \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \cos ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^2)^(-3/2),x]

[Out]

(Sqrt[a + b*Cos[x]^2]*EllipticE[Pi/2 + x, -(b/a)])/(a*(a + b)*Sqrt[1 + (b*Cos[x]^2)/a]) - (b*Cos[x]*Sin[x])/(a
*(a + b)*Sqrt[a + b*Cos[x]^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3256

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]/f)*EllipticE[e + f*x, -b/a], x] /
; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3257

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin
[e + f*x]^2/a)], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3263

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^
(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && N
eQ[a + b, 0] && LtQ[p, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.19, size = 75, normalized size = 0.96 \begin {gather*} \frac {2 (a+b) \sqrt {\frac {2 a+b+b \cos (2 x)}{a+b}} E\left (x\left |\frac {b}{a+b}\right .\right )-\sqrt {2} b \sin (2 x)}{2 a (a+b) \sqrt {2 a+b+b \cos (2 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^2)^(-3/2),x]

[Out]

(2*(a + b)*Sqrt[(2*a + b + b*Cos[2*x])/(a + b)]*EllipticE[x, b/(a + b)] - Sqrt[2]*b*Sin[2*x])/(2*a*(a + b)*Sqr
t[2*a + b + b*Cos[2*x]])

________________________________________________________________________________________

Maple [A]
time = 0.40, size = 73, normalized size = 0.94

method	result	size
default	\(-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {-\frac {b \left (\sin ^{2}\left (x \right )\right )}{a}+\frac {a +b}{a}}\, a \EllipticE \left (\cos \left (x \right ), \sqrt {-\frac {b}{a}}\right )+b \cos \left (x \right ) \left (\sin ^{2}\left (x \right )\right )}{a \left (a +b \right ) \sin \left (x \right ) \sqrt {a +b \left (\cos ^{2}\left (x \right )\right )}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-((sin(x)^2)^(1/2)*(-b/a*sin(x)^2+(a+b)/a)^(1/2)*a*EllipticE(cos(x),(-1/a*b)^(1/2))+b*cos(x)*sin(x)^2)/a/(a+b)
/sin(x)/(a+b*cos(x)^2)^(1/2)

________________________________________________________________________________________

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

[Out]

integrate((b*cos(x)^2 + a)^(-3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains complex when optimal does not.
time = 0.16, size = 775, normalized size = 9.94 \begin {gather*} -\frac {2 \, \sqrt {b \cos \left (x\right )^{2} + a} b^{3} \cos \left (x\right ) \sin \left (x\right ) + {\left (2 i \, a^{2} b + i \, a b^{2} + {\left (2 i \, a b^{2} + i \, b^{3}\right )} \cos \left (x\right )^{2} - 2 \, {\left (i \, b^{3} \cos \left (x\right )^{2} + i \, a b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, a^{2} b - i \, a b^{2} + {\left (-2 i \, a b^{2} - i \, b^{3}\right )} \cos \left (x\right )^{2} - 2 \, {\left (-i \, b^{3} \cos \left (x\right )^{2} - i \, a b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + 2 \, {\left (-2 i \, a^{3} - 3 i \, a^{2} b - i \, a b^{2} + {\left (-2 i \, a^{2} b - 3 i \, a b^{2} - i \, b^{3}\right )} \cos \left (x\right )^{2} + 2 \, {\left (-i \, a b^{2} \cos \left (x\right )^{2} - i \, a^{2} b\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + 2 \, {\left (2 i \, a^{3} + 3 i \, a^{2} b + i \, a b^{2} + {\left (2 i \, a^{2} b + 3 i \, a b^{2} + i \, b^{3}\right )} \cos \left (x\right )^{2} + 2 \, {\left (i \, a b^{2} \cos \left (x\right )^{2} + i \, a^{2} b\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{2 \, {\left (a^{3} b^{2} + a^{2} b^{3} + {\left (a^{2} b^{3} + a b^{4}\right )} \cos \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(b*cos(x)^2 + a)*b^3*cos(x)*sin(x) + (2*I*a^2*b + I*a*b^2 + (2*I*a*b^2 + I*b^3)*cos(x)^2 - 2*(I*b^
3*cos(x)^2 + I*a*b^2)*sqrt((a^2 + a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_e(
arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) + I*sin(x))), (8*a^2 + 8*a*b + b^2 + 4*(2*a*b + b
^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (-2*I*a^2*b - I*a*b^2 + (-2*I*a*b^2 - I*b^3)*cos(x)^2 - 2*(-I*b^3*cos(x)^2 -
I*a*b^2)*sqrt((a^2 + a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_e(arcsin(sqrt(
(2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) - I*sin(x))), (8*a^2 + 8*a*b + b^2 + 4*(2*a*b + b^2)*sqrt((a^
2 + a*b)/b^2))/b^2) + 2*(-2*I*a^3 - 3*I*a^2*b - I*a*b^2 + (-2*I*a^2*b - 3*I*a*b^2 - I*b^3)*cos(x)^2 + 2*(-I*a*
b^2*cos(x)^2 - I*a^2*b)*sqrt((a^2 + a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_
f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) + I*sin(x))), (8*a^2 + 8*a*b + b^2 + 4*(2*a*b +
b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + 2*(2*I*a^3 + 3*I*a^2*b + I*a*b^2 + (2*I*a^2*b + 3*I*a*b^2 + I*b^3)*cos(x)^
2 + 2*(I*a*b^2*cos(x)^2 + I*a^2*b)*sqrt((a^2 + a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b
)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) - I*sin(x))), (8*a^2 + 8*a*b + b^2 +
4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2))/(a^3*b^2 + a^2*b^3 + (a^2*b^3 + a*b^4)*cos(x)^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cos ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

integrate((b*cos(x)^2 + a)^(-3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)^2)^(3/2),x)

[Out]

int(1/(a + b*cos(x)^2)^(3/2), x)

________________________________________________________________________________________