3.384 \(\int \frac {1}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=109 \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (\sec (c+d x)+1)}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

[Out]

-(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))/a^2/d+2/3*(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))/a^2/d-1/3*sin(d*x+c)/d/cos(d*x+c)^(3/
2)/(a+a*sec(d*x+c))^2+sin(d*x+c)/a^2/d/(1+sec(d*x+c))/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4264, 3817, 4019, 3787, 3771, 2639, 2641} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (\sec (c+d x)+1)}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(EllipticE[(c + d*x)/2, 2]/(a^2*d)) + (2*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + Sin[c + d*x]/(a^2*d*Sqrt[Cos[
c + d*x]]*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2)

Rule 2639

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

Rule 2641

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

Rule 3771

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 3787

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 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {5 a}{2}+\frac {1}{2} a \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3 a^2}{2}-a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4}\\ &=\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{3 a^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2}-\frac {\int \sqrt {\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [C]  time = 6.18, size = 656, normalized size = 6.02 \[ -\frac {4 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin (c) \left (-\sqrt {\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{3 d \sqrt {\cot ^2(c)+1} (a \sec (c+d x)+a)^2}-\frac {i \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (\frac {2 e^{2 i d x} \sqrt {e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt {i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{3 i d \cos (c) \left (1+e^{2 i d x}\right )-3 d \sin (c) \left (-1+e^{2 i d x}\right )}-\frac {2 \sqrt {e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt {i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{d \sin (c) \left (-1+e^{2 i d x}\right )-i d \cos (c) \left (1+e^{2 i d x}\right )}\right )}{2 (a \sec (c+d x)+a)^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {2 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}-\frac {2 \tan \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {4 \csc (c)}{d}\right )}{\cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/2*I)*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4,
7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*S
in[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Co
s[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin
[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*
d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])
))/(a + a*Sec[c + d*x])^2 - (4*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Ar
cTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(
Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c
]^2]*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*((4*Csc[c])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/
2])/d - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(3*d) - (2*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(Cos[
c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.97, size = 257, normalized size = 2.36 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}{6 a^{2} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d*x+1/2*c)^6+4*(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))*cos(1/2*d*x+1/2*c)^3+6*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^3*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))-20*cos(1/2*d*x+1/2*c)^4+9*cos(1/2*d*x+1/2*c)^2-1)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
/cos(1/2*d*x+1/2*c)^3/sin(1/2*d*x+1/2*c)/(2*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 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )} + 2 \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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