3.383 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.749418, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4017, 4015, 3805, 3774, 203} \[ \frac{a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{5/2} (283 B+326 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{128 d}+\frac{a^2 (13 B+10 C) \sin (c+d x) \cos ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{40 d}+\frac{a^3 (157 B+170 C) \sin (c+d x) \cos ^2(c+d x)}{240 d \sqrt{a \sec (c+d x)+a}}+\frac{a^3 (283 B+326 C) \sin (c+d x) \cos (c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a B \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 4072

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

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 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 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 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])

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \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 B+10 C)+\frac{5}{2} a (B+2 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \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 B+170 C)+\frac{5}{4} a^2 (21 B+26 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \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 B+326 C)\right ) \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (283 B+326 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \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 B+326 C)\right ) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (283 B+326 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \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 B+326 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (283 B+326 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (a^3 (283 B+326 C)\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 B+326 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{128 d}+\frac{a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (283 B+326 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{a B \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 1.74728, size = 416, normalized size = 1.64 \[ \frac{a^2 \sin (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (15360 B \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},6,\frac{3}{2},1-\sec (c+d x)\right )+21504 C \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},5,\frac{3}{2},1-\sec (c+d x)\right )+11651 B \sqrt{1-\sec (c+d x)}+37029 B \cos (c+d x) \sqrt{1-\sec (c+d x)}+12653 B \cos (2 (c+d x)) \sqrt{1-\sec (c+d x)}+3818 B \cos (3 (c+d x)) \sqrt{1-\sec (c+d x)}+1002 B \cos (4 (c+d x)) \sqrt{1-\sec (c+d x)}+72 B \cos (5 (c+d x)) \sqrt{1-\sec (c+d x)}+25935 B \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )+9702 C \sqrt{1-\sec (c+d x)}+35658 C \cos (c+d x) \sqrt{1-\sec (c+d x)}+9786 C \cos (2 (c+d x)) \sqrt{1-\sec (c+d x)}+2436 C \cos (3 (c+d x)) \sqrt{1-\sec (c+d x)}+84 C \cos (4 (c+d x)) \sqrt{1-\sec (c+d x)}+28350 C \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{13440 d (\cos (c+d x)+1) \sqrt{1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(25935*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 28350*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 11651*B*Sqrt[1 - Sec
[c + d*x]] + 9702*C*Sqrt[1 - Sec[c + d*x]] + 37029*B*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 35658*C*Cos[c + d*x
]*Sqrt[1 - Sec[c + d*x]] + 12653*B*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 9786*C*Cos[2*(c + d*x)]*Sqrt[1 -
Sec[c + d*x]] + 3818*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 2436*C*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]
] + 1002*B*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 84*C*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 72*B*Cos[5
*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 21504*C*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c +
d*x]] + 15360*B*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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Maple [B]  time = 0.366, size = 947, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/61440/d*a^2*(4245*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+4890*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arct
anh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+16
980*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)
/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+19560*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+25470*B*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin
(d*x+c)*cos(d*x+c)^2*2^(1/2)+29340*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+16980*B*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+
c)*2^(1/2)+19560*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)*2^(1/2)+4245*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*2^(1/2)*ar
ctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+4890*C*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(9/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))
*sin(d*x+c)+12288*B*cos(d*x+c)^10+32256*B*cos(d*x+c)^9+15360*C*cos(d*x+c)^9+27904*B*cos(d*x+c)^8+43520*C*cos(d
*x+c)^8+18112*B*cos(d*x+c)^7+45440*C*cos(d*x+c)^7+45280*B*cos(d*x+c)^6+52160*C*cos(d*x+c)^6-135840*B*cos(d*x+c
)^5-156480*C*cos(d*x+c)^5)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^4

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.764589, size = 1227, normalized size = 4.83 \begin{align*} \left [\frac{15 \,{\left ({\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (283 \, B + 326 \, C\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 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \,{\left (29 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (283 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \,{\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \,{\left (283 \, B + 326 \, C\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 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (283 \, B + 326 \, C\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 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \,{\left (29 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (283 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \,{\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \,{\left (283 \, B + 326 \, C\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/3840*(15*((283*B + 326*C)*a^2*cos(d*x + c) + (283*B + 326*C)*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*B*a^2*cos(d*x + c)^5 + 48*(29*B + 10*C)*a^2*cos(d*x + c)^4 + 8*(283*B + 230*C)*a^2*cos(d*x + c)^3
+ 10*(283*B + 326*C)*a^2*cos(d*x + c)^2 + 15*(283*B + 326*C)*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*B + 326*C)*a^2*cos(d*x + c) + (283*B + 326*C)*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*B*a^2*c
os(d*x + c)^5 + 48*(29*B + 10*C)*a^2*cos(d*x + c)^4 + 8*(283*B + 230*C)*a^2*cos(d*x + c)^3 + 10*(283*B + 326*C
)*a^2*cos(d*x + c)^2 + 15*(283*B + 326*C)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+a*sec(d*x+c))**(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 8.22258, size = 1782, normalized size = 7.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/3840*(15*(283*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 326*C*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*B*sqrt(-a)*a^2*sgn
(cos(d*x + c)) + 326*C*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*sqrt(2)*(4245*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^18*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 4890*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a))^18*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 114615*(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)) - 132030*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 1298820*(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*sgn(cos(d*x + c)) + 1319880*(sqrt(-a)*tan(1/2*d*
x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 6176700*(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)) - 6888120*(sqrt(-a
)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 16394598*(
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)) + 183
52620*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a)*a^7*sgn(cos(d*x + c)
) - 14042770*(sqrt(-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)) - 15746180*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a^8*sgn(
cos(d*x + c)) + 4791060*(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)) + 5497320*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a
)*a^9*sgn(cos(d*x + c)) - 860300*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqr
t(-a)*a^10*sgn(cos(d*x + c)) - 959320*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*
C*sqrt(-a)*a^10*sgn(cos(d*x + c)) + 75885*(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)) + 84810*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^2*C*sqrt(-a)*a^11*sgn(cos(d*x + c)) - 2671*B*sqrt(-a)*a^12*sgn(cos(d*x + c)) - 2990*C*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