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

Optimal. Leaf size=261 \[ -\frac{b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (2 A+9 C)-4 a^4 C+3 A b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

[Out]

-((b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a
+ b)^(7/2)*d)) + (a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((3*A*b^4 - 4
*a^4*C + a^2*b^2*(2*A + 9*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + (a*(a^2*b^2*(2*A
- 5*C) + 2*a^4*C + b^4*(13*A + 18*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.672904, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4091, 4080, 4003, 12, 3831, 2659, 208} \[ -\frac{b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (2 A+9 C)-4 a^4 C+3 A b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a
+ b)^(7/2)*d)) + (a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((3*A*b^4 - 4
*a^4*C + a^2*b^2*(2*A + 9*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + (a*(a^2*b^2*(2*A
- 5*C) + 2*a^4*C + b^4*(13*A + 18*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 4091

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

Rule 4080

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[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f
*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e +
f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 4003

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) \left (-3 b \left (A b^2+a^2 C\right )+a \left (2 A b^2-\left (a^2-3 b^2\right ) C\right ) \sec (c+d x)+3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (-2 a b^2 \left (a^2 C-b^2 (5 A+6 C)\right )-b \left (a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\int -\frac{3 b^4 \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b^2 (A+2 C)+a^2 (4 A+3 C)\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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (4 a^2 A+A b^2+3 a^2 C+2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.27293, size = 221, normalized size = 0.85 \[ -\frac{\frac{2 \sin (c+d x) \left (6 b \left (9 a^2 b^2 (A+C)+a^4 (2 A+C)-A b^4\right ) \cos (c+d x)+a \left (\left (a^2 b^2 (10 A+11 C)+a^4 (6 A+4 C)-A b^4\right ) \cos (2 (c+d x))+a^2 b^2 (14 A+C)+a^4 (6 A+8 C)+b^4 (25 A+36 C)\right )\right )}{(a \cos (c+d x)+b)^3}+\frac{24 b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{24 d \left (b^2-a^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((24*b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2
] + (2*(6*b*(-(A*b^4) + 9*a^2*b^2*(A + C) + a^4*(2*A + C))*Cos[c + d*x] + a*(a^2*b^2*(14*A + C) + a^4*(6*A + 8
*C) + b^4*(25*A + 36*C) + (-(A*b^4) + a^4*(6*A + 4*C) + a^2*b^2*(10*A + 11*C))*Cos[2*(c + d*x)]))*Sin[c + d*x]
)/(b + a*Cos[c + d*x])^3)/(24*(-a^2 + b^2)^3*d)

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Maple [A]  time = 0.095, size = 374, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( 2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 2\,A{a}^{3}+2\,A{a}^{2}b+6\,Aa{b}^{2}+A{b}^{3}+2\,{a}^{3}C+3\,{a}^{2}bC+6\,Ca{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ \left ( a-b \right ) \left ({a}^{3}+3\,{a}^{2}b+3\,a{b}^{2}+{b}^{3} \right ) }}+2/3\,{\frac{ \left ( 3\,{a}^{2}A+7\,A{b}^{2}+{a}^{2}C+9\,{b}^{2}C \right ) a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}-1/2\,{\frac{ \left ( 2\,A{a}^{3}-2\,A{a}^{2}b+6\,Aa{b}^{2}-A{b}^{3}+2\,{a}^{3}C-3\,{a}^{2}bC+6\,Ca{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }} \right ) }-{\frac{b \left ( 4\,{a}^{2}A+A{b}^{2}+3\,{a}^{2}C+2\,{b}^{2}C \right ) }{{a}^{6}-3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}-{b}^{6}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(2*(-1/2*(2*A*a^3+2*A*a^2*b+6*A*a*b^2+A*b^3+2*C*a^3+3*C*a^2*b+6*C*a*b^2)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*t
an(1/2*d*x+1/2*c)^5+2/3*(3*A*a^2+7*A*b^2+C*a^2+9*C*b^2)*a/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3
-1/2*(2*A*a^3-2*A*a^2*b+6*A*a*b^2-A*b^3+2*C*a^3-3*C*a^2*b+6*C*a*b^2)/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d
*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3-b*(4*A*a^2+A*b^2+3*C*a^2+2*C*b^2)/(a^6-3*a^4*
b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.755258, size = 2475, normalized size = 9.48 \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)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[-1/12*(3*((4*A + 3*C)*a^2*b^4 + (A + 2*C)*b^6 + ((4*A + 3*C)*a^5*b + (A + 2*C)*a^3*b^3)*cos(d*x + c)^3 + 3*((
4*A + 3*C)*a^4*b^2 + (A + 2*C)*a^2*b^4)*cos(d*x + c)^2 + 3*((4*A + 3*C)*a^3*b^3 + (A + 2*C)*a*b^5)*cos(d*x + c
))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(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)) - 2*(2*C*a^7 + (2*A - 7*C)*a
^5*b^2 + (11*A + 23*C)*a^3*b^4 - (13*A + 18*C)*a*b^6 + (2*(3*A + 2*C)*a^7 + (4*A + 7*C)*a^5*b^2 - 11*(A + C)*a
^3*b^4 + A*a*b^6)*cos(d*x + c)^2 + 3*((2*A + C)*a^6*b + (7*A + 8*C)*a^4*b^3 - (10*A + 9*C)*a^2*b^5 + A*b^7)*co
s(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b -
4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^
8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d), -1/6*(3*((4*A + 3*C)*a^2
*b^4 + (A + 2*C)*b^6 + ((4*A + 3*C)*a^5*b + (A + 2*C)*a^3*b^3)*cos(d*x + c)^3 + 3*((4*A + 3*C)*a^4*b^2 + (A +
2*C)*a^2*b^4)*cos(d*x + c)^2 + 3*((4*A + 3*C)*a^3*b^3 + (A + 2*C)*a*b^5)*cos(d*x + c))*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))) - (2*C*a^7 + (2*A - 7*C)*a^5*b^2 + (11*A +
23*C)*a^3*b^4 - (13*A + 18*C)*a*b^6 + (2*(3*A + 2*C)*a^7 + (4*A + 7*C)*a^5*b^2 - 11*(A + C)*a^3*b^4 + A*a*b^6
)*cos(d*x + c)^2 + 3*((2*A + C)*a^6*b + (7*A + 8*C)*a^4*b^3 - (10*A + 9*C)*a^2*b^5 + A*b^7)*cos(d*x + c))*sin(
d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a
^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*co
s(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d)]

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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 ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29461, size = 936, normalized size = 3.59 \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)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(4*A*a^2*b + 3*C*a^2*b + A*b^3 + 2*C*b^3)*(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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt
(-a^2 + b^2)) + (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2
*c)^5 - 3*C*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*b^2*tan(1/2*d*x + 1/2
*c)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1
/2*c)^5 + 18*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^5*tan(1/2*d*x + 1/2*c)^3
- 4*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 16*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 32*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3
+ 28*A*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 36*C*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^5*tan(1/2*d*x + 1/2*c) + 6*C*a
^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b^2*tan(1
/2*d*x + 1/2*c) + 6*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 27*C*a^2*b^3*tan(1/2*
d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) + 18*C*a*b^4*tan(1/2*d*x + 1/2*c) - 3*A*b^5*tan(1/2*d*x + 1/2*c
))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3))/d