3.2.69 \(\int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx\) [169]

Optimal. Leaf size=155 \[ \frac {11 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}} \]

[Out]

11/4*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/d/a^(1/2)-2*A*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(
a-a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+5/4*A*sin(d*x+c)/d/(a-a*sec(d*x+c))^(1/2)+1/2*A*cos(d*x+c)*sin(d*x+c)
/d/(a-a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4107, 4005, 3859, 209, 3880} \begin {gather*} \frac {11 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (2*Sqrt[2]*A*ArcTan[(Sqrt[a]*Ta
n[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d) + (5*A*Sin[c + d*x])/(4*d*Sqrt[a - a*Sec[c + d*x]
]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a - a*Sec[c + d*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4107

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx &=\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-\frac {5 a A}{2}-\frac {3}{2} a A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a}\\ &=\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\frac {11 a^2 A}{4}+\frac {5}{4} a^2 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+(2 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx+\frac {(11 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{8 a}\\ &=\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+\frac {(11 A) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 d}-\frac {(4 A) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}\\ &=\frac {11 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.27, size = 332, normalized size = 2.14 \begin {gather*} \frac {A e^{-i (c+d x)} \left (7+6 e^{-i (c+d x)}+7 e^{i (c+d x)}+e^{-2 i (c+d x)}+6 e^{2 i (c+d x)}+e^{3 i (c+d x)}-11 i d \sqrt {1+e^{2 i (c+d x)}} x+11 \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+16 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}} \log \left (1-e^{i (c+d x)}\right )+11 \sqrt {1+e^{2 i (c+d x)}} \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-16 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}} \log \left (1+e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sec (c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+i \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{8 d \sqrt {a-a \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(7 + 6/E^(I*(c + d*x)) + 7*E^(I*(c + d*x)) + E^((-2*I)*(c + d*x)) + 6*E^((2*I)*(c + d*x)) + E^((3*I)*(c + d
*x)) - (11*I)*d*Sqrt[1 + E^((2*I)*(c + d*x))]*x + 11*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] +
16*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]*Log[1 - E^(I*(c + d*x))] + 11*Sqrt[1 + E^((2*I)*(c + d*x))]*Log[1 + S
qrt[1 + E^((2*I)*(c + d*x))]] - 16*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]*Log[1 + E^(I*(c + d*x)) + Sqrt[2]*Sqr
t[1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c + d*x)/2])/(8*d*E^(I*
(c + d*x))*Sqrt[a - a*Sec[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(366\) vs. \(2(130)=260\).
time = 8.88, size = 367, normalized size = 2.37

method	result	size
default	\(-\frac {A \left (-1+\cos \left (d x +c \right )\right )^{2} \left (16 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}\, \cos \left (d x +c \right )-6 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+16 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}-27 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )-48 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-15 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-48 \sqrt {2}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-66 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )-66 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )\right ) \sqrt {2}}{24 d \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (-1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{3}}\)	\(367\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/24*A/d*(-1+cos(d*x+c))^2*(16*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)*cos(d*x+c)-6*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^3+16*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)-27*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^2-4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)-48*2^(1/2)*cos(
d*x+c)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-15*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)-48*2^(1/
2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-66*cos(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*2^(1/2))-66*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)))/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(
a*(-1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^3*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.80, size = 462, normalized size = 2.98 \begin {gather*} \left [\frac {8 \, \sqrt {2} A a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 11 \, A \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, a d \sin \left (d x + c\right )}, \frac {8 \, \sqrt {2} A \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 11 \, A \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, a d \sin \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(8*sqrt(2)*A*a*sqrt(-1/a)*log(-(2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d
*x + c))*sqrt(-1/a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) - 11*
A*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) - (2*a*cos(
d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 2*(2*A*cos(d*x + c)^3 + 7*A*cos(d*x + c)^2 + 5*A*cos(
d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c)), 1/4*(8*sqrt(2)*A*sqrt(a)*arctan(sqrt(2)
*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) - 11*A*sqrt(a)*arct
an(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) - (2*A*cos(d*x +
c)^3 + 7*A*cos(d*x + c)^2 + 5*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} A \left (\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.06, size = 141, normalized size = 0.91 \begin {gather*} \frac {\frac {8 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {11 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {2} {\left (3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 10 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - 11*A*arctan(1/2*sqrt(2)*sqrt(a*t
an(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - sqrt(2)*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 10*sqrt(a*ta
n(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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