3.514 \(\int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=95 \[ \frac {3 F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{d}+\frac {4 \Pi \left (2;\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d} \]

[Out]

3/7*(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)+4/7*(
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)-(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),2/7*14^(1/2))/d*7^(1/2)+(3+4*cos(d*x+c))
^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.25, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2796, 3060, 2653, 3002, 2661, 2805} \[ \frac {3 F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{d}+\frac {4 \Pi \left (2;\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2653

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

Rule 2661

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

Rule 2796

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n)/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/
((m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*c*(m + 1) + b*d*n
+ (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(m + n + 2)*Sin[e + f*x]^2, x], x], 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] && LtQ[m, -1] && LtQ[0, n, 1] && Inte
gersQ[2*m, 2*n]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 3002

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 3060

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*} \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx &=\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{d}+\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)}{d}-\frac {1}{4} \int \frac {(-8-6 \cos (c+d x)) \sec (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx-\frac {1}{2} \int \sqrt {3+4 \cos (c+d x)} \, dx\\ &=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{d}+\frac {3}{2} \int \frac {1}{\sqrt {3+4 \cos (c+d x)}} \, dx+2 \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 )}{d}+\frac {3 F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}+\frac {4 \Pi \left (2;\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{\sqrt {7} d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [C]  time = 1.12, size = 157, normalized size = 1.65 \[ \frac {6 \sqrt {7} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {8}{7}\right )+21 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)+\frac {i \sqrt {7} \sin (c+d x) \left (-12 F\left (i \sinh ^{-1}\left (\sqrt {4 \cos (c+d x)+3}\right )|-\frac {1}{7}\right )+21 E\left (i \sinh ^{-1}\left (\sqrt {4 \cos (c+d x)+3}\right )|-\frac {1}{7}\right )-8 \Pi \left (-\frac {1}{3};i \sinh ^{-1}\left (\sqrt {4 \cos (c+d x)+3}\right )|-\frac {1}{7}\right )\right )}{\sqrt {\sin ^2(c+d x)}}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.80, size = 350, normalized size = 3.68 \[ -\frac {\sqrt {-\left (-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+\frac {3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-8*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)*(-8*sin(1/2*d*x+1/2*c)^4+7*s
in(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*cos(1/2*d*x+1/2*c)^2+
1)^(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))+(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-8*cos(1/2*d*x+1/2*c)^2+1)^(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*cos(1/2*d*x+1/2*c)^2+1)^(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {4\,\cos \left (c+d\,x\right )+3}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {4 \cos {\left (c + d x \right )} + 3} \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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