3.560 \(\int \frac{\sec ^3(c+d x)}{\sqrt{3-4 \cos (c+d x)}} \, dx\)

Optimal. Leaf size=140 \[ \frac{F\left (\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 \sqrt{7} d}-\frac{\sqrt{7} E\left (\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 d}-\frac{\sqrt{7} \Pi \left (2;\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x) \sec (c+d x)}{6 d} \]

[Out]

-(Sqrt[7]*EllipticE[(c + Pi + d*x)/2, 8/7])/(3*d) + EllipticF[(c + Pi + d*x)/2, 8/7]/(3*Sqrt[7]*d) - (Sqrt[7]*
EllipticPi[2, (c + Pi + d*x)/2, 8/7])/(3*d) + (Sqrt[3 - 4*Cos[c + d*x]]*Tan[c + d*x])/(3*d) + (Sqrt[3 - 4*Cos[
c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.366455, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2802, 3055, 3059, 2654, 3002, 2662, 2806} \[ \frac{F\left (\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 \sqrt{7} d}-\frac{\sqrt{7} E\left (\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 d}-\frac{\sqrt{7} \Pi \left (2;\frac{1}{2} (c+d x+\pi )|\frac{8}{7}\right )}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x) \sec (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[7]*EllipticE[(c + Pi + d*x)/2, 8/7])/(3*d) + EllipticF[(c + Pi + d*x)/2, 8/7]/(3*Sqrt[7]*d) - (Sqrt[7]*
EllipticPi[2, (c + Pi + d*x)/2, 8/7])/(3*d) + (Sqrt[3 - 4*Cos[c + d*x]]*Tan[c + d*x])/(3*d) + (Sqrt[3 - 4*Cos[
c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(6*d)

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*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] + Dis
t[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[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + 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, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2654

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 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 2662

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 2806

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]

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{\sqrt{3-4 \cos (c+d x)}} \, dx &=\frac{\sqrt{3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} \int \frac{\left (6+3 \cos (c+d x)-2 \cos ^2(c+d x)\right ) \sec ^2(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{\sqrt{3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{18} \int \frac{\left (21-6 \cos (c+d x)+12 \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{\sqrt{3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{72} \int \frac{(84+12 \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+\pi +d x)|\frac{8}{7}\right )}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} \int \frac{1}{\sqrt{3-4 \cos (c+d x)}} \, dx+\frac{7}{6} \int \frac{\sec (c+d x)}{\sqrt{3-4 \cos (c+d x)}} \, dx\\ &=-\frac{\sqrt{7} E\left (\frac{1}{2} (c+\pi +d x)|\frac{8}{7}\right )}{3 d}+\frac{F\left (\frac{1}{2} (c+\pi +d x)|\frac{8}{7}\right )}{3 \sqrt{7} d}-\frac{\sqrt{7} \Pi \left (2;\frac{1}{2} (c+\pi +d x)|\frac{8}{7}\right )}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac{\sqrt{3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [C]  time = 1.74908, size = 236, normalized size = 1.69 \[ \frac{-\frac{4 \sqrt{4 \cos (c+d x)-3} F\left (\left .\frac{1}{2} (c+d x)\right |8\right )}{\sqrt{3-4 \cos (c+d x)}}+\frac{18 \sqrt{4 \cos (c+d x)-3} \Pi \left (2;\left .\frac{1}{2} (c+d x)\right |8\right )}{\sqrt{3-4 \cos (c+d x)}}+\sqrt{3-4 \cos (c+d x)} (2 \cos (c+d x)+1) \tan (c+d x) \sec (c+d x)-\frac{2 i \sin (c+d x) \left (-12 F\left (i \sinh ^{-1}\left (\sqrt{3-4 \cos (c+d x)}\right )|-\frac{1}{7}\right )+21 E\left (i \sinh ^{-1}\left (\sqrt{3-4 \cos (c+d x)}\right )|-\frac{1}{7}\right )-8 \Pi \left (-\frac{1}{3};i \sinh ^{-1}\left (\sqrt{3-4 \cos (c+d x)}\right )|-\frac{1}{7}\right )\right )}{3 \sqrt{7} \sqrt{\sin ^2(c+d x)}}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*Sqrt[-3 + 4*Cos[c + d*x]]*EllipticF[(c + d*x)/2, 8])/Sqrt[3 - 4*Cos[c + d*x]] + (18*Sqrt[-3 + 4*Cos[c + d
*x]]*EllipticPi[2, (c + d*x)/2, 8])/Sqrt[3 - 4*Cos[c + d*x]] - (((2*I)/3)*(21*EllipticE[I*ArcSinh[Sqrt[3 - 4*C
os[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]]*(1 +
2*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d)

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Maple [B]  time = 3.833, size = 408, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/3*cos(1/2*d*x+1/2*c)*(8*sin(1/2*d*x+1/2*c)^4-sin
(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(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-sin(1/2*d
*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1
/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2/7*14^(1/2))-1/3*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))-1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)/(8
*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2)))/sin(1/2*d*x+1
/2*c)/(-8*cos(1/2*d*x+1/2*c)^2+7)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{3}}{\sqrt{-4 \, \cos \left (d x + c\right ) + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{3}}{4 \, \cos \left (d x + c\right ) - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\sqrt{3 - 4 \cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{3}}{\sqrt{-4 \, \cos \left (d x + c\right ) + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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