3.172 \(\int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {5 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {7 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

[Out]

5*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d-A*sin(d*x+c)/d/(a-a*sec(d*x+c))^(3/2)-7/2*A*ar
ctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+2*A*sin(d*x+c)/a/d/(a-a*sec(d*x+
c))^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac {5 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {7 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

(5*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(3/2)*d) - (7*A*ArcTan[(Sqrt[a]*Tan[c + d*x])
/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (A*Sin[c + d*x])/(d*(a - a*Sec[c + d*x])^(3/2)) +
(2*A*Sin[c + d*x])/(a*d*Sqrt[a - a*Sec[c + d*x]])

Rule 203

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

Rule 3774

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 3795

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 3920

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 4020

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

Rule 4022

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 (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx &=-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) (4 a A+3 a A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {-5 a^2 A-2 a^2 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^3}\\ &=-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}+\frac {(5 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{2 a^2}+\frac {(7 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}+\frac {(5 A) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a d}-\frac {(7 A) \operatorname {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 )}{a d}\\ &=\frac {5 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {7 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}-\frac {A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {2 A \sin (c+d x)}{a d \sqrt {a-a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.63, size = 361, normalized size = 2.47 \[ A \left (\frac {\sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (-\frac {2 \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}+\frac {2 \sin \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{d}+\frac {2 \cos \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{d}-\frac {2 \cos \left (\frac {3 c}{2}\right ) \cos \left (\frac {3 d x}{2}\right )}{d}-\frac {2 \cot \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {2 \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}+\frac {\sqrt {2} e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (-5 \sinh ^{-1}\left (e^{i (c+d x)}\right )+7 \sqrt {2} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-5 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{d (a-a \sec (c+d x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

A*((Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-5*ArcSinh[E^(I*(c
+ d*x))] + 7*Sqrt[2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - 5*ArcTanh[Sqrt[1
+ E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sin[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d*x])
^(3/2)) + (Sec[c + d*x]^2*((2*Cos[c/2]*Cos[(d*x)/2])/d - (2*Cos[(3*c)/2]*Cos[(3*d*x)/2])/d - (2*Cot[c/2]*Csc[c
/2 + (d*x)/2])/d + (2*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/d - (2*Sin[c/2]*Sin[(d*x)/2])/d + (2*Sin[(3*
c)/2]*Sin[(3*d*x)/2])/d)*Sin[c/2 + (d*x)/2]^3)/(a - a*Sec[c + d*x])^(3/2))

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fricas [A]  time = 0.48, size = 526, normalized size = 3.60 \[ \left [-\frac {7 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {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 (3 \, a \cos \left (d x + c\right ) + a\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 ) + 10 \, {\left (A \cos \left (d x + c\right ) - A\right )} \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 ) + 4 \, {\left (A \cos \left (d x + c\right )^{3} - A \cos \left (d x + c\right )^{2} - 2 \, 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 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {7 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\right )} \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 ) - 10 \, {\left (A \cos \left (d x + c\right ) - A\right )} \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 ) - 2 \, {\left (A \cos \left (d x + c\right )^{3} - A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(7*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a
*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin
(d*x + c) + 10*(A*cos(d*x + c) - 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) + 4*(A*cos(d*x + c)^3
- A*cos(d*x + c)^2 - 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)
*sin(d*x + c)), 1/2*(7*sqrt(2)*(A*cos(d*x + c) - 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) - 10*(A*cos(d*x + c) - A)*sqrt(a)*arctan(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 - A*cos(d*x
+ c)^2 - 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c)
)]

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giac [B]  time = 2.21, size = 261, normalized size = 1.79 \[ -\frac {\frac {7 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {10 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 4 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a}{{\left ({\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} + 3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 2 \, a^{2}\right )} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(7*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*
sgn(tan(1/2*d*x + 1/2*c))) - 10*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(
tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - (3*sqrt(2)*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 4
*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A*a)/(((a*tan(1/2*d*x + 1/2*c)^2 - a)^2 + 3*(a*tan(1/2*d*x + 1/2*c
)^2 - a)*a + 2*a^2)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))))/d

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maple [B]  time = 1.64, size = 462, normalized size = 3.16 \[ \frac {A \left (-1+\cos \left (d x +c \right )\right )^{3} \left (-3 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-6 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right ) \sqrt {2}-7 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-3 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}+3 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+21 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+7 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}+30 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sqrt {2}-21 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}-6 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-30 \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}}{3 d \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\frac {a \left (-1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*A/d*(-1+cos(d*x+c))^3*(-3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*2^(1/2)-6*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(5/2)*cos(d*x+c)*2^(1/2)-7*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*2^(1/2)-3*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*2^(1/2)+21*arctan(1/(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2*2^(1/2)-2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*2^(1
/2)+7*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)+30*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))
*cos(d*x+c)^2+5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*2^(1/2)-21*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2))*2^(1/2)-6*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)-30*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)))^(3/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(3/2)/sin(d*x+c)^5*2^(1/
2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((A*sec(d*x + c) + A)*cos(d*x + c)/(-a*sec(d*x + c) + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(3/2),x)

[Out]

int((cos(c + d*x)*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(3/2),x)

[Out]

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

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