3.691 \(\int \frac{\sec ^4(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=381 \[ -\frac{a \left (a^2 b^2 (2 A-21 C)+12 a^4 C-b^4 (5 A-6 C)\right ) \tan (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}+\frac{\left (C \left (12 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\left (7 a^2 b^2 C-4 a^4 C+3 A b^4\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (A-10 C)+6 a^4 C-b^4 (4 A-C)\right ) \tan (c+d x) \sec (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2} \]

[Out]

((2*A*b^2 + (12*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/(2*b^5*d) - (a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^
4*(A - 4*C) + 12*a^6*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*
d) - (a*(a^2*b^2*(2*A - 21*C) - b^4*(5*A - 6*C) + 12*a^4*C)*Tan[c + d*x])/(2*b^4*(a^2 - b^2)^2*d) + ((a^2*b^2*
(A - 10*C) - b^4*(4*A - C) + 6*a^4*C)*Sec[c + d*x]*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - ((A*b^2 + a^2*C)*Se
c[c + d*x]^3*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((3*A*b^4 - 4*a^4*C + 7*a^2*b^2*C)*Sec
[c + d*x]^2*Tan[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.62036, antiderivative size = 381, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4099, 4098, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{a \left (a^2 b^2 (2 A-21 C)+12 a^4 C-b^4 (5 A-6 C)\right ) \tan (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}+\frac{\left (C \left (12 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\left (7 a^2 b^2 C-4 a^4 C+3 A b^4\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (A-10 C)+6 a^4 C-b^4 (4 A-C)\right ) \tan (c+d x) \sec (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

((2*A*b^2 + (12*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/(2*b^5*d) - (a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^
4*(A - 4*C) + 12*a^6*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*
d) - (a*(a^2*b^2*(2*A - 21*C) - b^4*(5*A - 6*C) + 12*a^4*C)*Tan[c + d*x])/(2*b^4*(a^2 - b^2)^2*d) + ((a^2*b^2*
(A - 10*C) - b^4*(4*A - C) + 6*a^4*C)*Sec[c + d*x]*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - ((A*b^2 + a^2*C)*Se
c[c + d*x]^3*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((3*A*b^4 - 4*a^4*C + 7*a^2*b^2*C)*Sec
[c + d*x]^2*Tan[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4099

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

Rule 4098

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

Rule 4092

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

Rule 4082

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec ^3(c+d x) \left (3 \left (A b^2+a^2 C\right )-2 a b (A+C) \sec (c+d x)-2 \left (A b^2+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\sec ^2(c+d x) \left (2 \left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right )-a b \left (3 A b^2-\left (a^2-4 b^2\right ) C\right ) \sec (c+d x)+2 \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (2 a \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right )-2 b \left (2 a^4 C-b^4 (2 A+C)-a^2 b^2 (A+4 C)\right ) \sec (c+d x)-2 a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \tan (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (2 a b \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right )+2 \left (a^2-b^2\right )^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{4 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \tan (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (a \left (6 A b^6+a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^5 \left (a^2-b^2\right )^2}+\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \int \sec (c+d x) \, dx}{2 b^5}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \tan (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (a \left (6 A b^6+a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^6 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \tan (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (a \left (6 A b^6+a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^6 \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{a \left (2 a^4 A b^2-5 a^2 A b^4+6 A b^6+12 a^6 C-29 a^4 b^2 C+20 a^2 b^4 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^5 (a+b)^{5/2} d}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \tan (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.14997, size = 559, normalized size = 1.47 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{2 a^2 b^2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{(b-a) (a+b)}+\frac{2 a^2 b \left (a^2 b^2 (9 C-2 A)-6 a^4 C+5 A b^4\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}-2 \left (C \left (12 a^2+b^2\right )+2 A b^2\right ) (a \cos (c+d x)+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (C \left (12 a^2+b^2\right )+2 A b^2\right ) (a \cos (c+d x)+b)^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{b^2 C (a \cos (c+d x)+b)^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C (a \cos (c+d x)+b)^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{12 a b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{12 a b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{2 b^5 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((4*a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^4*(A
- 4*C) + 12*a^6*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(
5/2) - 2*(2*A*b^2 + (12*a^2 + b^2)*C)*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A
*b^2 + (12*a^2 + b^2)*C)*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^2*C*(b + a*Cos[c
+ d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - (12*a*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]) - (b^2*C*(b + a*Cos[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (1
2*a*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (2*a^2*b^2*(A*b^2 + a
^2*C)*Sin[c + d*x])/((-a + b)*(a + b)) + (2*a^2*b*(5*A*b^4 - 6*a^4*C + a^2*b^2*(-2*A + 9*C))*(b + a*Cos[c + d*
x])*Sin[c + d*x])/((a - b)^2*(a + b)^2)))/(2*b^5*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.109, size = 1547, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

1/2/d*C/b^3/(tan(1/2*d*x+1/2*c)+1)+1/2/d*C/b^3/(tan(1/2*d*x+1/2*c)-1)^2-1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*A-1/2
/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d*C/b^3/(tan(1/2*d*x+1/2*c)-1)-1/2/d*C/b^3/(tan(1/2*d*x+1/2*c)+1)^2+1/d/
b^3*ln(tan(1/2*d*x+1/2*c)+1)*A+1/2/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*C+29/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a
-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-20/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b)
)^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-2/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^
(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+5/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2
)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-12/d*a^7/b^5/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)
*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*
b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a
-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A-6/d*a*b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2
*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-
b)^2*tan(1/2*d*x+1/2*c)*A-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/
2*d*x+1/2*c)*A-6/d*a^6/b^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2
*c)*C-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)
^3*A-1/d*a^5/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C+2/d*
a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+10/
d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C+6/d*a^6/b^4
/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-1/d*a^5/b^
3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-10/d*a^4/
b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+6/d/b^5
*ln(tan(1/2*d*x+1/2*c)+1)*a^2*C+3/d*C/b^4/(tan(1/2*d*x+1/2*c)-1)*a+3/d*C/b^4/(tan(1/2*d*x+1/2*c)+1)*a-6/d/b^5*
ln(tan(1/2*d*x+1/2*c)-1)*a^2*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 78.1789, size = 4635, normalized size = 12.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((12*C*a^9 + (2*A - 29*C)*a^7*b^2 - 5*(A - 4*C)*a^5*b^4 + 6*A*a^3*b^6)*cos(d*x + c)^4 + 2*(12*C*a^8*b +
(2*A - 29*C)*a^6*b^3 - 5*(A - 4*C)*a^4*b^5 + 6*A*a^2*b^7)*cos(d*x + c)^3 + (12*C*a^7*b^2 + (2*A - 29*C)*a^5*b^
4 - 5*(A - 4*C)*a^3*b^6 + 6*A*a*b^8)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*c
os(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b
*cos(d*x + c) + b^2)) + ((12*C*a^10 + (2*A - 35*C)*a^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A - 3*C)*a^4*b^6 -
(2*A + C)*a^2*b^8)*cos(d*x + c)^4 + 2*(12*C*a^9*b + (2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5*b^5 + 3*(2*A - 3
*C)*a^3*b^7 - (2*A + C)*a*b^9)*cos(d*x + c)^3 + (12*C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A - 11*C)*a^4*b^6
+ 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)*b^10)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((12*C*a^10 + (2*A - 35*C)*a
^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A - 3*C)*a^4*b^6 - (2*A + C)*a^2*b^8)*cos(d*x + c)^4 + 2*(12*C*a^9*b +
(2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5*b^5 + 3*(2*A - 3*C)*a^3*b^7 - (2*A + C)*a*b^9)*cos(d*x + c)^3 + (12*
C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A - 11*C)*a^4*b^6 + 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)*b^10)*cos(d*x +
c)^2)*log(-sin(d*x + c) + 1) + 2*(C*a^6*b^4 - 3*C*a^4*b^6 + 3*C*a^2*b^8 - C*b^10 - (12*C*a^9*b + (2*A - 33*C)*
a^7*b^3 - (7*A - 27*C)*a^5*b^5 + (5*A - 6*C)*a^3*b^7)*cos(d*x + c)^3 - (18*C*a^8*b^2 + (3*A - 50*C)*a^6*b^4 -
(9*A - 43*C)*a^4*b^6 + (6*A - 11*C)*a^2*b^8)*cos(d*x + c)^2 - 4*(C*a^7*b^3 - 3*C*a^5*b^5 + 3*C*a^3*b^7 - C*a*b
^9)*cos(d*x + c))*sin(d*x + c))/((a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*d*cos(d*x + c)^4 + 2*(a^7*b^6 -
3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*d*cos(d*x + c)^3 + (a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*d*cos(d*x + c)^2
), -1/4*(2*((12*C*a^9 + (2*A - 29*C)*a^7*b^2 - 5*(A - 4*C)*a^5*b^4 + 6*A*a^3*b^6)*cos(d*x + c)^4 + 2*(12*C*a^8
*b + (2*A - 29*C)*a^6*b^3 - 5*(A - 4*C)*a^4*b^5 + 6*A*a^2*b^7)*cos(d*x + c)^3 + (12*C*a^7*b^2 + (2*A - 29*C)*a
^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 6*A*a*b^8)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x
+ c) + a)/((a^2 - b^2)*sin(d*x + c))) - ((12*C*a^10 + (2*A - 35*C)*a^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A
- 3*C)*a^4*b^6 - (2*A + C)*a^2*b^8)*cos(d*x + c)^4 + 2*(12*C*a^9*b + (2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5
*b^5 + 3*(2*A - 3*C)*a^3*b^7 - (2*A + C)*a*b^9)*cos(d*x + c)^3 + (12*C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A
- 11*C)*a^4*b^6 + 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)*b^10)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((12*C*a^10
+ (2*A - 35*C)*a^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A - 3*C)*a^4*b^6 - (2*A + C)*a^2*b^8)*cos(d*x + c)^4 +
2*(12*C*a^9*b + (2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5*b^5 + 3*(2*A - 3*C)*a^3*b^7 - (2*A + C)*a*b^9)*cos(
d*x + c)^3 + (12*C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A - 11*C)*a^4*b^6 + 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)
*b^10)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(C*a^6*b^4 - 3*C*a^4*b^6 + 3*C*a^2*b^8 - C*b^10 - (12*C*a^9*
b + (2*A - 33*C)*a^7*b^3 - (7*A - 27*C)*a^5*b^5 + (5*A - 6*C)*a^3*b^7)*cos(d*x + c)^3 - (18*C*a^8*b^2 + (3*A -
50*C)*a^6*b^4 - (9*A - 43*C)*a^4*b^6 + (6*A - 11*C)*a^2*b^8)*cos(d*x + c)^2 - 4*(C*a^7*b^3 - 3*C*a^5*b^5 + 3*
C*a^3*b^7 - C*a*b^9)*cos(d*x + c))*sin(d*x + c))/((a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*d*cos(d*x + c)^
4 + 2*(a^7*b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*d*cos(d*x + c)^3 + (a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)
*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.34949, size = 1602, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(12*C*a^7 + 2*A*a^5*b^2 - 29*C*a^5*b^2 - 5*A*a^3*b^4 + 20*C*a^3*b^4 + 6*A*a*b^6)*(pi*floor(1/2*(d*x +
c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((
a^4*b^5 - 2*a^2*b^7 + b^9)*sqrt(-a^2 + b^2)) - 2*(12*C*a^7*tan(1/2*d*x + 1/2*c)^7 - 18*C*a^6*b*tan(1/2*d*x + 1
/2*c)^7 + 2*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 - 17*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^4*b^3*tan(1/2*d*x +
1/2*c)^7 + 33*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^7 - 5*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^7 - 2*C*a^3*b^4*tan(1/2*d*x
+ 1/2*c)^7 + 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 - 13*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 + 4*C*a*b^6*tan(1/2*d*x
+ 1/2*c)^7 + C*b^7*tan(1/2*d*x + 1/2*c)^7 - 36*C*a^7*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^6*b*tan(1/2*d*x + 1/2*c)
^5 - 6*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 + 67*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^4*b^3*tan(1/2*d*x + 1/2*
c)^5 - 29*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^5 + 15*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^5 - 26*C*a^3*b^4*tan(1/2*d*x +
1/2*c)^5 - 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 5*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 4*C*a*b^6*tan(1/2*d*x + 1
/2*c)^5 + 3*C*b^7*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^7*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^6*b*tan(1/2*d*x + 1/2*c)^3
+ 6*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 - 67*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*b^3*tan(1/2*d*x + 1/2*c)
^3 - 29*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 26*C*a^3*b^4*tan(1/2*d*x + 1/
2*c)^3 - 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 + 5*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*b^6*tan(1/2*d*x + 1/2
*c)^3 + 3*C*b^7*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^7*tan(1/2*d*x + 1/2*c) - 18*C*a^6*b*tan(1/2*d*x + 1/2*c) - 2*A
*a^5*b^2*tan(1/2*d*x + 1/2*c) + 17*C*a^5*b^2*tan(1/2*d*x + 1/2*c) - 3*A*a^4*b^3*tan(1/2*d*x + 1/2*c) + 33*C*a^
4*b^3*tan(1/2*d*x + 1/2*c) + 5*A*a^3*b^4*tan(1/2*d*x + 1/2*c) + 2*C*a^3*b^4*tan(1/2*d*x + 1/2*c) + 6*A*a^2*b^5
*tan(1/2*d*x + 1/2*c) - 13*C*a^2*b^5*tan(1/2*d*x + 1/2*c) - 4*C*a*b^6*tan(1/2*d*x + 1/2*c) + C*b^7*tan(1/2*d*x
+ 1/2*c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x
+ 1/2*c)^2 + a + b)^2) - (12*C*a^2 + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^5 + (12*C*a^2 + 2*
A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^5)/d