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

Optimal. Leaf size=254 \[ \frac {a^{5/2} (283 A+326 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{240 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (283 A+326 B) \sin (c+d x) \cos (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{40 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

1/128*a^(5/2)*(283*A+326*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/5*a*A*cos(d*x+c)^4*(a+a*sec(
d*x+c))^(3/2)*sin(d*x+c)/d+1/128*a^3*(283*A+326*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/192*a^3*(283*A+326*B)
*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/240*a^3*(157*A+170*B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x
+c))^(1/2)+1/40*a^2*(13*A+10*B)*cos(d*x+c)^3*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.65, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4017, 4015, 3805, 3774, 203} \[ \frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^{5/2} (283 A+326 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^2 (13 A+10 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{40 d}+\frac {a^3 (157 A+170 B) \sin (c+d x) \cos ^2(c+d x)}{240 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (283 A+326 B) \sin (c+d x) \cos (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(283*A + 326*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^3*(283*A + 326*
B)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(283*A + 326*B)*Cos[c + d*x]*Sin[c + d*x])/(192*d*Sqr
t[a + a*Sec[c + d*x]]) + (a^3*(157*A + 170*B)*Cos[c + d*x]^2*Sin[c + d*x])/(240*d*Sqrt[a + a*Sec[c + d*x]]) +
(a^2*(13*A + 10*B)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(40*d) + (a*A*Cos[c + d*x]^4*(a + a*S
ec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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 3805

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

Rule 4015

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

Rule 4017

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (13 A+10 B)+\frac {5}{2} a (A+2 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (157 A+170 B)+\frac {5}{4} a^2 (21 A+26 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{96} \left (a^2 (283 A+326 B)\right ) \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (283 A+326 B) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{128} \left (a^2 (283 A+326 B)\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (283 A+326 B) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{256} \left (a^2 (283 A+326 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (283 A+326 B) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (a^3 (283 A+326 B)\right ) \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 )}{128 d}\\ &=\frac {a^{5/2} (283 A+326 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (283 A+326 B) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 A+170 B) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+10 B) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 1.82, size = 416, normalized size = 1.64 \[ \frac {a^2 \sin (c+d x) \sqrt {a (\sec (c+d x)+1)} \left (15360 A \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},6;\frac {3}{2};1-\sec (c+d x)\right )+11651 A \sqrt {1-\sec (c+d x)}+37029 A \cos (c+d x) \sqrt {1-\sec (c+d x)}+12653 A \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+3818 A \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+1002 A \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+72 A \cos (5 (c+d x)) \sqrt {1-\sec (c+d x)}+25935 A \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )+21504 B \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},5;\frac {3}{2};1-\sec (c+d x)\right )+9702 B \sqrt {1-\sec (c+d x)}+35658 B \cos (c+d x) \sqrt {1-\sec (c+d x)}+9786 B \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+2436 B \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+84 B \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+28350 B \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )\right )}{13440 d (\cos (c+d x)+1) \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(25935*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 28350*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 11651*A*Sqrt[1 - Sec
[c + d*x]] + 9702*B*Sqrt[1 - Sec[c + d*x]] + 37029*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 35658*B*Cos[c + d*x
]*Sqrt[1 - Sec[c + d*x]] + 12653*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 9786*B*Cos[2*(c + d*x)]*Sqrt[1 -
Sec[c + d*x]] + 3818*A*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 2436*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]
] + 1002*A*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 84*B*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 72*A*Cos[5
*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 21504*B*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c +
d*x]] + 15360*A*Hypergeometric2F1[1/2, 6, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(13440*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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fricas [A]  time = 0.74, size = 460, normalized size = 1.81 \[ \left [\frac {15 \, {\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (384 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3840 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (384 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1920 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3840*(15*((283*A + 326*B)*a^2*cos(d*x + c) + (283*A + 326*B)*a^2)*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*(384*A*a^2*cos(d*x + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8*(283*A + 230*B)*a^2*cos(d*x + c)^3
+ 10*(283*A + 326*B)*a^2*cos(d*x + c)^2 + 15*(283*A + 326*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/1920*(15*((283*A + 326*B)*a^2*cos(d*x + c) + (283*A + 326*B)*a
^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (384*A*a^2*c
os(d*x + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8*(283*A + 230*B)*a^2*cos(d*x + c)^3 + 10*(283*A + 326*B
)*a^2*cos(d*x + c)^2 + 15*(283*A + 326*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c) + d)]

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giac [B]  time = 4.18, size = 1377, normalized size = 5.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3840*(15*(283*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 326*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*ta
n(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 15*(283*A*sqrt(-a)*a^2*sgn
(cos(d*x + c)) + 326*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/
2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(4245*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^18*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 4890*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 114615*sqrt(2)*(sqrt(-a)*tan(1/2*d*x
+ 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 132030*sqrt(2)*(sqrt(-a
)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 1298820*sq
rt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*sqrt(-a)*a^5*sgn(cos(d*x + c)
) + 1319880*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a^5*sg
n(cos(d*x + c)) - 6176700*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*s
qrt(-a)*a^6*sgn(cos(d*x + c)) - 6888120*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a))^12*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 16394598*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/
2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 18352620*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 14042770*sqrt(2)*(sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 15746180*sqrt(2)*(sq
rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 479106
0*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^9*sgn(cos(d*x +
c)) + 5497320*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^9*
sgn(cos(d*x + c)) - 860300*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*s
qrt(-a)*a^10*sgn(cos(d*x + c)) - 959320*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a))^4*B*sqrt(-a)*a^10*sgn(cos(d*x + c)) + 75885*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d
*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^11*sgn(cos(d*x + c)) + 84810*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^11*sgn(cos(d*x + c)) - 2671*sqrt(2)*A*sqrt(-a)*a^12*sgn(cos(d*x
+ c)) - 2990*sqrt(2)*B*sqrt(-a)*a^12*sgn(cos(d*x + c)))/((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)^5)/d

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maple [B]  time = 1.61, size = 947, normalized size = 3.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/61440/d*(4245*A*sin(d*x+c)*cos(d*x+c)^4*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)))^(9/2)*2^(1/2)+4890*B*sin(d*x+c)*cos(d*x+c)^4*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)))^(9/2)*2^(1/2)+16980*
A*sin(d*x+c)*cos(d*x+c)^3*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)))^(9/2)*2^(1/2)+19560*B*sin(d*x+c)*cos(d*x+c)^3*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)))^(9/2)*2^(1/2)+25470*A*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)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(9/2)*2^(1/2)+29340*B*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)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)+16980*A*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)))^(9/2)*2
^(1/2)+19560*B*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)))^(9/2)*2^(1/2)+4245*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)))^(9/2)*sin(d*x+c)+4890*B*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)))^(9/2)*sin
(d*x+c)+12288*A*cos(d*x+c)^10+32256*A*cos(d*x+c)^9+15360*B*cos(d*x+c)^9+27904*A*cos(d*x+c)^8+43520*B*cos(d*x+c
)^8+18112*A*cos(d*x+c)^7+45440*B*cos(d*x+c)^7+45280*A*cos(d*x+c)^6+52160*B*cos(d*x+c)^6-135840*A*cos(d*x+c)^5-
156480*B*cos(d*x+c)^5)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^4*a^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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