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

Optimal. Leaf size=135 \[ \frac{3 F\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{\sqrt{7} d}-\frac{\sqrt{7} E\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{3 d}+\frac{5 \Pi \left (2;\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{3 \sqrt{7} d}+\frac{\sqrt{4 \cos (c+d x)+3} \tan (c+d x)}{3 d}+\frac{\sqrt{4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

-(Sqrt[7]*EllipticE[(c + d*x)/2, 8/7])/(3*d) + (3*EllipticF[(c + d*x)/2, 8/7])/(Sqrt[7]*d) + (5*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) + (Sqrt[3 + 4*Cos[c + d*x]]*S
ec[c + d*x]*Tan[c + d*x])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.363618, antiderivative size = 135, 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 = {2796, 3055, 3059, 2653, 3002, 2661, 2805} \[ \frac{3 F\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{\sqrt{7} d}-\frac{\sqrt{7} E\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{3 d}+\frac{5 \Pi \left (2;\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{3 \sqrt{7} d}+\frac{\sqrt{4 \cos (c+d x)+3} \tan (c+d x)}{3 d}+\frac{\sqrt{4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[7]*EllipticE[(c + d*x)/2, 8/7])/(3*d) + (3*EllipticF[(c + d*x)/2, 8/7])/(Sqrt[7]*d) + (5*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) + (Sqrt[3 + 4*Cos[c + d*x]]*S
ec[c + d*x]*Tan[c + d*x])/(2*d)

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \sqrt{3+4 \cos (c+d x)} \sec ^3(c+d x) \, dx &=\frac{\sqrt{3+4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \frac{\left (2+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)}{2 d}+\frac{1}{6} \int \frac{\left (5+6 \cos (c+d x)-4 \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)}{2 d}-\frac{1}{24} \int \frac{(-20-36 \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{\sqrt{3+4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d}+\frac{5}{6} \int \frac{\sec (c+d x)}{\sqrt{3+4 \cos (c+d x)}} \, dx+\frac{3}{2} \int \frac{1}{\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{3 F\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{\sqrt{7} d}+\frac{5 \Pi \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}+\frac{\sqrt{3+4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [C]  time = 1.24727, size = 194, normalized size = 1.44 \[ \frac{\frac{12 F\left (\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{\sqrt{7}}+\frac{6 \Pi \left (2;\frac{1}{2} (c+d x)|\frac{8}{7}\right )}{\sqrt{7}}+(2 \cos (c+d x)+3) \sqrt{4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)+\frac{2 i \sin (c+d x) \left (-12 F\left (i \sinh ^{-1}\left (\sqrt{4 \cos (c+d x)+3}\right )|-\frac{1}{7}\right )+21 E\left (i \sinh ^{-1}\left (\sqrt{4 \cos (c+d x)+3}\right )|-\frac{1}{7}\right )-8 \Pi \left (-\frac{1}{3};i \sinh ^{-1}\left (\sqrt{4 \cos (c+d x)+3}\right )|-\frac{1}{7}\right )\right )}{3 \sqrt{7} \sqrt{\sin ^2(c+d x)}}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*EllipticF[(c + d*x)/2, 8/7])/Sqrt[7] + (6*EllipticPi[2, (c + d*x)/2, 8/7])/Sqrt[7] + (((2*I)/3)*(21*Ellip
ticE[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]) + (3
+ 2*Cos[c + d*x])*Sqrt[3 + 4*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(6*d)

________________________________________________________________________________________

Maple [B]  time = 3.497, size = 408, normalized size = 3. \begin{align*} -{\frac{1}{d}\sqrt{- \left ( -8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -{\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-2}}-{\frac{2}{3}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1}}+3\,{\frac{\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,2\,\sqrt{2} \right ) }{\sqrt{-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}+{\frac{1}{3}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{-8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,2\,\sqrt{2} \right ){\frac{1}{\sqrt{-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}}-{\frac{5}{3}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{-8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticPi} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,2,2\,\sqrt{2} \right ){\frac{1}{\sqrt{-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-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)^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)+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))+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)
*EllipticE(cos(1/2*d*x+1/2*c),2*2^(1/2))-5/3*(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)*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

________________________________________________________________________________________

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

________________________________________________________________________________________

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

________________________________________________________________________________________

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

________________________________________________________________________________________

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