3.165 \(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=207 \[ -\frac {(5 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-(5*A-2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))
^(5/2)-1/16*(15*A-7*B)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/32*(115*A-43*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2
^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/16*(35*A-11*B)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(5 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

-(((5*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4
*d*(a + a*Sec[c + d*x])^(5/2)) - ((15*A - 7*B)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((35*A - 11
*B)*Sin[c + d*x])/(16*a^2*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+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\cos (c+d x) \left (a (5 A-B)-\frac {5}{2} a (A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (\frac {1}{2} a^2 (35 A-11 B)-\frac {3}{4} a^2 (15 A-7 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {-4 a^3 (5 A-2 B)+\frac {1}{4} a^3 (35 A-11 B) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(115 A-43 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}-\frac {(5 A-2 B) \int \sqrt {a+a \sec (c+d x)} \, dx}{2 a^3}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(115 A-43 B) \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 )}{16 a^2 d}+\frac {(5 A-2 B) \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^2 d}\\ &=-\frac {(5 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A-11 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 27.31, size = 12012, normalized size = 58.03 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

Result too large to show

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fricas [A]  time = 8.20, size = 739, normalized size = 3.57 \[ \left [\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 115 \, A - 43 \, B\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 32 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (16 \, A \cos \left (d x + c\right )^{3} + 5 \, {\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 115 \, A - 43 \, B\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 ) - 32 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B\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 ) - 2 \, {\left (16 \, A \cos \left (d x + c\right )^{3} + 5 \, {\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((115*A - 43*B)*cos(d*x + c)^3 + 3*(115*A - 43*B)*cos(d*x + c)^2 + 3*(115*A - 43*B)*cos(d*x + c
) + 115*A - 43*B)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d
*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 32*((5*A - 2*B)*
cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)*cos(d*x + c) + 5*A - 2*B)*sqrt(-a)*log((2*a*cos(
d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a
)/(cos(d*x + c) + 1)) + 4*(16*A*cos(d*x + c)^3 + 5*(11*A - 3*B)*cos(d*x + c)^2 + (35*A - 11*B)*cos(d*x + c))*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*
cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((115*A - 43*B)*cos(d*x + c)^3 + 3*(115*A - 43*B)*cos(d*x + c)^2 + 3*(11
5*A - 43*B)*cos(d*x + c) + 115*A - 43*B)*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))) - 32*((5*A - 2*B)*cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)
*cos(d*x + c) + 5*A - 2*B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*
x + c))) - 2*(16*A*cos(d*x + c)^3 + 5*(11*A - 3*B)*cos(d*x + c)^2 + (35*A - 11*B)*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c)
+ a^3*d)]

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giac [B]  time = 4.67, size = 499, normalized size = 2.41 \[ \frac {2 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\sqrt {2} {\left (21 \, A a^{5} - 13 \, B a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (115 \, A - 43 \, B\right )} \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {32 \, {\left (5 \, A - 2 \, B\right )} \log \left (\frac {{\left | -562949953421312 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 1125899906842624 \, \sqrt {2} {\left | a \right |} + 1688849860263936 \, a \right |}}{{\left | -562949953421312 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 1125899906842624 \, \sqrt {2} {\left | a \right |} + 1688849860263936 \, a \right |}}\right )}{\sqrt {-a} a {\left | a \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {128 \, \sqrt {2} {\left (3 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A - A a\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} \sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(tan(1/2
*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(21*A*a^5 - 13*B*a^5)/(a^8*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2
*c) + sqrt(2)*(115*A - 43*B)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqr
t(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 32*(5*A - 2*B)*log(abs(-562949953421312*(sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 1125899906842624*sqrt(2)*abs(a) + 1688849860263936*a)/abs(-56
2949953421312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 1125899906842624*sqrt(
2)*abs(a) + 1688849860263936*a))/(sqrt(-a)*a*abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 128*sqrt(2)*(3*(sqrt(-a
)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A - A*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
^2*a + a^2)*sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.63, size = 1065, normalized size = 5.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/32/d*(-1+cos(d*x+c))^2*(80*A*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-32*B*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1
/2)+115*A*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+160*A*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
anh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-43*B*sin(d*x+c)*ln(-(-(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos
(d*x+c)^2-64*B*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+230*A*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+80*A*2^(1/2)*arctanh(1/
2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)-32*A*cos(d*x+c)^4-86*B*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos
(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-32*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(
1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)+115*A*ln(-(-(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)-78*A*cos(d*x+c)^3-43*B*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+30*B*cos(d*x+c)^3+40*A*cos(d*x+c)^2-8*B*cos(d*x+c)^2+70*A*cos(d*
x+c)-22*B*cos(d*x+c))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^5/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*cos(d*x + c)/(a*sec(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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