3.374 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=282 \[ \frac{2 a^2 (13 B+16 C) \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{143 d}+\frac{2 a^3 (299 B+280 C) \tan (c+d x) \sec ^4(c+d x)}{1287 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (4615 B+4184 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}-\frac{8 a^2 (4615 B+4184 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{45045 d}+\frac{4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a (4615 B+4184 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{13 d} \]

[Out]

(4*a^3*(4615*B + 4184*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(4615*B + 4184*C)*Sec[c + d*
x]^3*Tan[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(299*B + 280*C)*Sec[c + d*x]^4*Tan[c + d*x])/(12
87*d*Sqrt[a + a*Sec[c + d*x]]) - (8*a^2*(4615*B + 4184*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(45045*d) + (
2*a^2*(13*B + 16*C)*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(143*d) + (4*a*(4615*B + 4184*C)*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(15015*d) + (2*a*C*Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x
])/(13*d)

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Rubi [A]  time = 0.839559, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4072, 4018, 4016, 3803, 3800, 4001, 3792} \[ \frac{2 a^2 (13 B+16 C) \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{143 d}+\frac{2 a^3 (299 B+280 C) \tan (c+d x) \sec ^4(c+d x)}{1287 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (4615 B+4184 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}-\frac{8 a^2 (4615 B+4184 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{45045 d}+\frac{4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a (4615 B+4184 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(4*a^3*(4615*B + 4184*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(4615*B + 4184*C)*Sec[c + d*
x]^3*Tan[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(299*B + 280*C)*Sec[c + d*x]^4*Tan[c + d*x])/(12
87*d*Sqrt[a + a*Sec[c + d*x]]) - (8*a^2*(4615*B + 4184*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(45045*d) + (
2*a^2*(13*B + 16*C)*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(143*d) + (4*a*(4615*B + 4184*C)*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(15015*d) + (2*a*C*Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x
])/(13*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 4018

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

Rule 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, 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 3803

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

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

Rule 3792

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

Rubi steps

\begin{align*} \int \sec ^3(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 \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{2}{13} \int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (13 B+8 C)+\frac{1}{2} a (13 B+16 C) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{4}{143} \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (247 B+216 C)+\frac{1}{4} a^2 (299 B+280 C) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{\left (a^2 (4615 B+4184 C)\right ) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{1287}\\ &=\frac{2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{\left (2 a^2 (4615 B+4184 C)\right ) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{3003}\\ &=\frac{2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{(4 a (4615 B+4184 C)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{15015}\\ &=\frac{2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt{a+a \sec (c+d x)}}-\frac{8 a^2 (4615 B+4184 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac{\left (2 a^2 (4615 B+4184 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{6435}\\ &=\frac{4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt{a+a \sec (c+d x)}}-\frac{8 a^2 (4615 B+4184 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac{2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac{4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}\\ \end{align*}

Mathematica [A]  time = 0.478743, size = 131, normalized size = 0.46 \[ \frac{2 a^3 \tan (c+d x) \left (315 (13 B+38 C) \sec ^5(c+d x)+35 (416 B+523 C) \sec ^4(c+d x)+5 (4615 B+4184 C) \sec ^3(c+d x)+6 (4615 B+4184 C) \sec ^2(c+d x)+8 (4615 B+4184 C) \sec (c+d x)+73840 B+3465 C \sec ^6(c+d x)+66944 C\right )}{45045 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^3*(73840*B + 66944*C + 8*(4615*B + 4184*C)*Sec[c + d*x] + 6*(4615*B + 4184*C)*Sec[c + d*x]^2 + 5*(4615*B
+ 4184*C)*Sec[c + d*x]^3 + 35*(416*B + 523*C)*Sec[c + d*x]^4 + 315*(13*B + 38*C)*Sec[c + d*x]^5 + 3465*C*Sec[c
+ d*x]^6)*Tan[c + d*x])/(45045*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.365, size = 185, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 73840\,B \left ( \cos \left ( dx+c \right ) \right ) ^{6}+66944\,C \left ( \cos \left ( dx+c \right ) \right ) ^{6}+36920\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+33472\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+27690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+25104\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+23075\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+20920\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+14560\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18305\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4095\,B\cos \left ( dx+c \right ) +11970\,C\cos \left ( dx+c \right ) +3465\,C \right ) }{45045\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/45045/d*a^2*(-1+cos(d*x+c))*(73840*B*cos(d*x+c)^6+66944*C*cos(d*x+c)^6+36920*B*cos(d*x+c)^5+33472*C*cos(d*x
+c)^5+27690*B*cos(d*x+c)^4+25104*C*cos(d*x+c)^4+23075*B*cos(d*x+c)^3+20920*C*cos(d*x+c)^3+14560*B*cos(d*x+c)^2
+18305*C*cos(d*x+c)^2+4095*B*cos(d*x+c)+11970*C*cos(d*x+c)+3465*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x
+c)^6/sin(d*x+c)

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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(sec(d*x+c)^3*(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 = 0.538267, size = 479, normalized size = 1.7 \begin{align*} \frac{2 \,{\left (16 \,{\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 8 \,{\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 6 \,{\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \,{\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \,{\left (416 \, B + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \,{\left (13 \, B + 38 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/45045*(16*(4615*B + 4184*C)*a^2*cos(d*x + c)^6 + 8*(4615*B + 4184*C)*a^2*cos(d*x + c)^5 + 6*(4615*B + 4184*C
)*a^2*cos(d*x + c)^4 + 5*(4615*B + 4184*C)*a^2*cos(d*x + c)^3 + 35*(416*B + 523*C)*a^2*cos(d*x + c)^2 + 315*(1
3*B + 38*C)*a^2*cos(d*x + c) + 3465*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c
)^7 + d*cos(d*x + c)^6)

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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(sec(d*x+c)**3*(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 [A]  time = 5.25256, size = 486, normalized size = 1.72 \begin{align*} \frac{8 \,{\left (45045 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 45045 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (150150 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 120120 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (300300 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 294294 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (356070 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 310596 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (232375 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 212069 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (21125 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 19279 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (1625 \, \sqrt{2} B a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, \sqrt{2} C a^{9} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{45045 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/45045*(45045*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 45045*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - (150150*sqrt(2)*B*a^9
*sgn(cos(d*x + c)) + 120120*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - (300300*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 294294
*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - (356070*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 310596*sqrt(2)*C*a^9*sgn(cos(d*x
+ c)) - (232375*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 212069*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - 4*(21125*sqrt(2)*B*
a^9*sgn(cos(d*x + c)) + 19279*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - 2*(1625*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 1483
*sqrt(2)*C*a^9*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(
1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2
*c)^2 - a)^6*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)