\(\int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [716]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b d}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d} \]

[Out]

2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d
*x+c)^(1/2)/b/d-2*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(
d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/d+2*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d
*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a+b)/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3317, 3937, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 d (a+b)}-\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d} \]

[In]

Int[1/((a + b*Cos[c + d*x])*Sec[c + d*x]^(3/2)),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*d) - (2*a*Sqrt[Cos[c + d*x]]*EllipticF[
(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*d) + (2*a^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2]*Sqrt[Sec[c + d*x]])/(b^2*(a + b)*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3934

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3937

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[b^2/(a^2
*d^2), Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a - b*Csc[e + f*x])/Sqrt[d*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx \\ & = \frac {\int \frac {b-a \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{b^2}+\frac {a^2 \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{b^2} \\ & = -\frac {a \int \sqrt {\sec (c+d x)} \, dx}{b^2}+\frac {\int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{b}+\frac {\left (a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^2} \\ & = \frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d}-\frac {\left (a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{b^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{b} \\ & = \frac {2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b d}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 19.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cot (c+d x) \left (-b \sec ^{\frac {3}{2}}(c+d x)-b \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+b \sec ^{\frac {7}{2}}(c+d x)+b \cos (2 (c+d x)) \sec ^{\frac {7}{2}}(c+d x)-2 b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 a \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{b^2 d} \]

[In]

Integrate[1/((a + b*Cos[c + d*x])*Sec[c + d*x]^(3/2)),x]

[Out]

(Cot[c + d*x]*(-(b*Sec[c + d*x]^(3/2)) - b*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + b*Sec[c + d*x]^(7/2) + b*Cos[
2*(c + d*x)]*Sec[c + d*x]^(7/2) - 2*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 2*b*El
lipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*a*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x
]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(b^2*d)

Maple [A] (verified)

Time = 3.78 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-a^{2} \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(227\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral(1/((a + b*cos(c + d*x))*sec(c + d*x)**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(1/((b*cos(d*x + c) + a)*sec(d*x + c)^(3/2)), x)

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(1/((b*cos(d*x + c) + a)*sec(d*x + c)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

[In]

int(1/((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))),x)

[Out]

int(1/((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))), x)