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

Optimal. Leaf size=266 \[ -\frac {(47 A+24 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac {(17 A+9 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a \sec (c+d x)+a}}+\frac {(5 A+3 C) \sin (c+d x) \cos ^2(c+d x)}{6 a d \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(13 A+6 C) \sin (c+d x) \cos (c+d x)}{12 a d \sqrt {a \sec (c+d x)+a}} \]

[Out]

-1/8*(47*A+24*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/2*(A+C)*cos(d*x+c)^2*sin(d*x+c)
/d/(a+a*sec(d*x+c))^(3/2)+1/4*(17*A+9*C)*arctan(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)+3/8*(7*A+4*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/12*(13*A+6*C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*
sec(d*x+c))^(1/2)+1/6*(5*A+3*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.78, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4085, 4022, 3920, 3774, 203, 3795} \[ -\frac {(47 A+24 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac {(17 A+9 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a \sec (c+d x)+a}}+\frac {(5 A+3 C) \sin (c+d x) \cos ^2(c+d x)}{6 a d \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(13 A+6 C) \sin (c+d x) \cos (c+d x)}{12 a d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((47*A + 24*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*a^(3/2)*d) + ((17*A + 9*C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Cos[c + d*x]^2*Si
n[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (3*(7*A + 4*C)*Sin[c + d*x])/(8*a*d*Sqrt[a + a*Sec[c + d*x]]) -
((13*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A + 3*C)*Cos[c + d*x]^2*Sin[
c + d*x])/(6*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 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]

Rule 4085

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\cos ^3(c+d x) \left (-a (5 A+3 C)+\frac {1}{2} a (7 A+3 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (13 A+6 C)-\frac {5}{2} a^2 (5 A+3 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{6 a^3}\\ &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(13 A+6 C) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-\frac {9}{2} a^3 (7 A+4 C)+\frac {3}{2} a^3 (13 A+6 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{12 a^4}\\ &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+6 C) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\frac {3}{4} a^4 (47 A+24 C)-\frac {9}{4} a^4 (7 A+4 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+6 C) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}+\frac {(17 A+9 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}-\frac {(47 A+24 C) \int \sqrt {a+a \sec (c+d x)} \, dx}{16 a^2}\\ &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+6 C) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}-\frac {(17 A+9 C) \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 )}{2 a d}+\frac {(47 A+24 C) \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 )}{8 a d}\\ &=-\frac {(47 A+24 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 a^{3/2} d}+\frac {(17 A+9 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {3 (7 A+4 C) \sin (c+d x)}{8 a d \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+6 C) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.11, size = 204, normalized size = 0.77 \[ -\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (\sin ^2\left (\frac {1}{2} (c+d x)\right ) ((43 A+24 C) \cos (c+d x)-3 A \cos (2 (c+d x))+2 A \cos (3 (c+d x))+60 A+36 C)-3 (47 A+24 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )+6 \sqrt {2} (17 A+9 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right )\right )}{12 a d (\cos (c+d x)-1) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((-3*(47*A + 24*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + 6*Sqrt[2
]*(17*A + 9*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + (60*A + 36
*C + (43*A + 24*C)*Cos[c + d*x] - 3*A*Cos[2*(c + d*x)] + 2*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]^2)*Tan[(c + d*
x)/2])/(a*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 5.80, size = 668, normalized size = 2.51 \[ \left [-\frac {6 \, \sqrt {2} {\left ({\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 17 \, A + 9 \, C\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 ) + 3 \, {\left ({\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right ) + 47 \, A + 24 \, C\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 ) - 2 \, {\left (8 \, A \cos \left (d x + c\right )^{4} - 6 \, A \cos \left (d x + c\right )^{3} + {\left (37 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (7 \, A + 4 \, C\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 )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {6 \, \sqrt {2} {\left ({\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 17 \, A + 9 \, C\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 ) - 3 \, {\left ({\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right ) + 47 \, A + 24 \, C\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 ) - {\left (8 \, A \cos \left (d x + c\right )^{4} - 6 \, A \cos \left (d x + c\right )^{3} + {\left (37 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (7 \, A + 4 \, C\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 )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(6*sqrt(2)*((17*A + 9*C)*cos(d*x + c)^2 + 2*(17*A + 9*C)*cos(d*x + c) + 17*A + 9*C)*sqrt(-a)*log((2*sqr
t(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)) + 3*((47*A + 24*C)*cos(d*x + c)^2 + 2*(47*A + 24*C)*cos(
d*x + c) + 47*A + 24*C)*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)) - 2*(8*A*cos(d*x + c)^4 - 6*A*cos(d*x + c)
^3 + (37*A + 24*C)*cos(d*x + c)^2 + 9*(7*A + 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*
x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/24*(6*sqrt(2)*((17*A + 9*C)*cos(d*x + c)^2 +
2*(17*A + 9*C)*cos(d*x + c) + 17*A + 9*C)*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))) - 3*((47*A + 24*C)*cos(d*x + c)^2 + 2*(47*A + 24*C)*cos(d*x + c) + 47*A + 24*
C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (8*A*cos(d*x
+ c)^4 - 6*A*cos(d*x + c)^3 + (37*A + 24*C)*cos(d*x + c)^2 + 9*(7*A + 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]

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giac [B]  time = 10.03, size = 850, normalized size = 3.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(6*sqrt(2)*(17*A + 9*C)*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*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 3*(47*A + 24*C)*log(abs(-1947111321950560360698936123457536*(sqrt(-
a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 3894222643901120721397872246915072*sqrt(2)*
abs(a) + 5841333965851681082096808370372608*a)/abs(-1947111321950560360698936123457536*(sqrt(-a)*tan(1/2*d*x +
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 3894222643901120721397872246915072*sqrt(2)*abs(a) + 5841333
965851681082096808370372608*a))/(sqrt(-a)*abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 12*(sqrt(2)*A*a*sgn(tan(1/
2*d*x + 1/2*c)^2 - 1) + sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1
/2*d*x + 1/2*c)/a^3 - 4*sqrt(2)*(339*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*
A + 72*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C - 3165*(sqrt(-a)*tan(1/2*d*x
+ 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a - 888*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^8*C*a + 9198*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^
2 + 3024*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^2 - 4938*(sqrt(-a)*tan(1/
2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^3 - 1776*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*
tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a^3 + 975*(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^4 + 360*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^4 - 73*A*a^5 - 24
*C*a^5)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-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)^3*sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.90, size = 1414, normalized size = 5.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192/d*(-1+cos(d*x+c))*(-141*A*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*sin(d*x+c)-72*C*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(5/2)*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))*sin(d*
x+c)-204*A*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))-423*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(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)))^(5/2)-108*C*cos(d*x+c)^3*sin(d*
x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/si
n(d*x+c))-216*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/co
s(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-612*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)))^(5/2)*sin(d*x+c)*cos(d*x+c)^2-423*A*2^(1/2)*sin(d*
x+c)*cos(d*x+c)*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)))^(5/2)-324*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)))^(5/2)*sin(d*x+c)*cos(d*x+c)^2-216*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)*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)))^(5/2)-612*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)))^(5/2
)*sin(d*x+c)*cos(d*x+c)-141*A*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)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)-324*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)))^(5/2)*sin(d*x+c)*cos(d*x+c)-72*C*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)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*s
in(d*x+c)+64*A*cos(d*x+c)^7-204*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)))^(5/2)*sin(d*x+c)-108*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-co
s(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)-112*A*cos(d*x+c)^6+344*A*cos(d*x+c)^5+
192*C*cos(d*x+c)^5+208*A*cos(d*x+c)^4+96*C*cos(d*x+c)^4-504*A*cos(d*x+c)^3-288*C*cos(d*x+c)^3)*(a*(1+cos(d*x+c
))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^3/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)

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