\(\int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx\) [552]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 \sqrt {7} d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d} \]

[Out]

1/7*(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),2/7*14^(1/2))/d*7^(1/2)-4/21*
(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),2,2/7*14^(1/2))/d*7^(1/2)-1/3*(c
os(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),2/7*14^(1/2))/d*7^(1/2)+1/3*(3+4*co
s(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2881, 3139, 2732, 3081, 2740, 2884} \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}-\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 \sqrt {7} d}+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{3 d} \]

[In]

Int[Sec[c + d*x]^2/Sqrt[3 + 4*Cos[c + d*x]],x]

[Out]

-1/3*(Sqrt[7]*EllipticE[(c + d*x)/2, 8/7])/d + EllipticF[(c + d*x)/2, 8/7]/(Sqrt[7]*d) - (4*EllipticPi[2, (c +
d*x)/2, 8/7])/(3*Sqrt[7]*d) + (Sqrt[3 + 4*Cos[c + d*x]]*Tan[c + d*x])/(3*d)

Rule 2732

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

Rule 2740

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

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

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 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3139

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]], x], x] - Dist[1/(b*d), Int[Sim
p[a*c*C - A*b*d + (b*c*C + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /;
FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} \int \frac {\left (-2-2 \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = \frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d}-\frac {1}{12} \int \frac {(8-6 \cos (c+d x)) \sec (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx-\frac {1}{6} \int \sqrt {3+4 \cos (c+d x)} \, dx \\ & = -\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{2} \int \frac {1}{\sqrt {3+4 \cos (c+d x)}} \, dx-\frac {2}{3} \int \frac {\sec (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = -\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 \sqrt {7} d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7}}+\frac {i \left (21 E\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{3 \sqrt {7} \sqrt {\sin ^2(c+d x)}}+\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d} \]

[In]

Integrate[Sec[c + d*x]^2/Sqrt[3 + 4*Cos[c + d*x]],x]

[Out]

((-6*EllipticPi[2, (c + d*x)/2, 8/7])/Sqrt[7] + ((I/3)*(21*EllipticE[I*ArcSinh[Sqrt[3 + 4*Cos[c + d*x]]], -1/7
] - 12*EllipticF[I*ArcSinh[Sqrt[3 + 4*Cos[c + d*x]]], -1/7] - 8*EllipticPi[-1/3, I*ArcSinh[Sqrt[3 + 4*Cos[c +
d*x]]], -1/7])*Sin[c + d*x])/(Sqrt[7]*Sqrt[Sin[c + d*x]^2]) + Sqrt[3 + 4*Cos[c + d*x]]*Tan[c + d*x])/(3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(166)=332\).

Time = 2.53 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.47

method	result	size
default	\(-\frac {\sqrt {-\left (1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{3 \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )}{3 \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(350\)

[In]

int(sec(d*x+c)^2/(3+4*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(-(1-8*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/3*cos(1/2*d*x+1/2*c)*(-8*sin(1/2*d*x+1/2*c)^4+7*
sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/2*c)^2)
^(1/2)/(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2*2^(1/2))+1/3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticE(cos(1/2*d*x+1/2*c),2*2^(1/2))+4/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/2*c)^2)^(1/2)/(
-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,2*2^(1/2)))/sin(1/2*d*x+
1/2*c)/(8*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^2/(3+4*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sec(d*x + c)^2/sqrt(4*cos(d*x + c) + 3), x)

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {4 \cos {\left (c + d x \right )} + 3}}\, dx \]

[In]

integrate(sec(d*x+c)**2/(3+4*cos(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**2/sqrt(4*cos(c + d*x) + 3), x)

Maxima [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^2/(3+4*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/sqrt(4*cos(d*x + c) + 3), x)

Giac [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^2/(3+4*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/sqrt(4*cos(d*x + c) + 3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {4\,\cos \left (c+d\,x\right )+3}} \,d x \]

[In]

int(1/(cos(c + d*x)^2*(4*cos(c + d*x) + 3)^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^2*(4*cos(c + d*x) + 3)^(1/2)), x)