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

Optimal. Leaf size=271 \[ -\frac{a \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{\left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C+3 a^4 C-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\left (3 a^2 C+2 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b^2 d \left (a^2-b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.86113, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4099, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{a \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{\left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C+3 a^4 C-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\left (3 a^2 C+2 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b^2 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b^2 + (6*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (2*a*(a^2*A*b^2 - 2*A*b^4 + 3*a^4*C - 4*a^2*b^
2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^4*(a + b)^(3/2)*d) - (a*(A*b^2 + 3*
a^2*C - 2*b^2*C)*Tan[c + d*x])/(b^3*(a^2 - b^2)*d) + ((2*A*b^2 + 3*a^2*C - b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(
2*b^2*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^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 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 ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-a b (A+C) \sec (c+d x)-\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a \left (2 A b^2+\left (3 a^2-b^2\right ) C\right )+b \left (2 A b^2+\left (a^2+b^2\right ) C\right ) \sec (c+d x)+2 a \left (A b^2+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a b \left (b^2 (2 A-C)+3 a^2 C\right )-\left (a^2-b^2\right ) \left (2 A b^2+6 a^2 C+b^2 C\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{a \left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (a \left (2 A b^4-a^2 b^2 (A-4 C)-3 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^4 \left (a^2-b^2\right )}+\frac{\left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \int \sec (c+d x) \, dx}{2 b^4}\\ &=\frac{\left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{a \left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (a \left (2 A b^4-a^2 b^2 (A-4 C)-3 a^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=\frac{\left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{a \left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 a \left (2 A b^4-a^2 b^2 (A-4 C)-3 a^4 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^5 \left (a^2-b^2\right ) d}\\ &=\frac{\left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-2 A b^4+3 a^4 C-4 a^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)^{3/2} b^4 (a+b)^{3/2} d}-\frac{a \left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.79693, size = 461, normalized size = 1.7 \[ \frac{(a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{4 a^2 b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(b-a) (a+b)}-2 \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 a \left (a^2 b^2 (A-4 C)+3 a^4 C-2 A b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b^2 C (a \cos (c+d x)+b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C (a \cos (c+d x)+b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{8 a b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{8 a b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{2 b^4 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*((8*a*(-2*A*b^4 + a^2*b^2*(A - 4*C) + 3*a^4*C)*ArcTanh[((-a + b)*
Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2 - b^2)^(3/2) - 2*(2*A*b^2 + (6*a^2 + b^2)*C)*(b
+ a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 + (6*a^2 + b^2)*C)*(b + a*Cos[c + d*x]
)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^2*C*(b + a*Cos[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
)^2 - (8*a*b*C*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (b^2*C*(b + a*Co
s[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (8*a*b*C*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c
 + d*x)/2] + Sin[(c + d*x)/2]) + (4*a^2*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((-a + b)*(a + b))))/(2*b^4*d*(A + 2*C
 + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.104, size = 646, normalized size = 2.4 \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)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)

[Out]

2/d*a^2/b/(a^2-b^2)*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)*A+2/d*a^4/b^3/(a^2-
b^2)*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)*C-2/d*a^3/b^2/(a+b)/(a-b)/((a+b)*(
a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+4/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arct
anh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-6/d*a^5/b^4/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*
tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+8/d*a^3/b^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*
x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/2/d*C/b^2/(tan(1/2*d*x+1/2*c)+1)^2+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*A+3/d/b^
4*ln(tan(1/2*d*x+1/2*c)+1)*a^2*C+1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*C+2/d*C/b^3/(tan(1/2*d*x+1/2*c)+1)*a+1/2/d
*C/b^2/(tan(1/2*d*x+1/2*c)+1)+1/2/d*C/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*A-3/d/b^4*
ln(tan(1/2*d*x+1/2*c)-1)*a^2*C-1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*C+2/d*C/b^3/(tan(1/2*d*x+1/2*c)-1)*a+1/2/d*C
/b^2/(tan(1/2*d*x+1/2*c)-1)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 29.9701, size = 2553, normalized size = 9.42 \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)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(2*((3*C*a^6 + (A - 4*C)*a^4*b^2 - 2*A*a^2*b^4)*cos(d*x + c)^3 + (3*C*a^5*b + (A - 4*C)*a^3*b^3 - 2*A*a*b
^5)*cos(d*x + c)^2)*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)) + ((6*C*a^7
 + (2*A - 11*C)*a^5*b^2 - 4*(A - C)*a^3*b^4 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b + (2*A - 11*C)*a^4*
b^3 - 4*(A - C)*a^2*b^5 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((6*C*a^7 + (2*A - 11*C)*a^5*
b^2 - 4*(A - C)*a^3*b^4 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b + (2*A - 11*C)*a^4*b^3 - 4*(A - C)*a^2*
b^5 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7 - 2*(3*C*a^6*
b + (A - 5*C)*a^4*b^3 - (A - 2*C)*a^2*b^5)*cos(d*x + c)^2 - 3*(C*a^5*b^2 - 2*C*a^3*b^4 + C*a*b^6)*cos(d*x + c)
)*sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2
), -1/4*(4*((3*C*a^6 + (A - 4*C)*a^4*b^2 - 2*A*a^2*b^4)*cos(d*x + c)^3 + (3*C*a^5*b + (A - 4*C)*a^3*b^3 - 2*A*
a*b^5)*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
))) - ((6*C*a^7 + (2*A - 11*C)*a^5*b^2 - 4*(A - C)*a^3*b^4 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b + (2
*A - 11*C)*a^4*b^3 - 4*(A - C)*a^2*b^5 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((6*C*a^7 + (2
*A - 11*C)*a^5*b^2 - 4*(A - C)*a^3*b^4 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b + (2*A - 11*C)*a^4*b^3 -
 4*(A - C)*a^2*b^5 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(C*a^4*b^3 - 2*C*a^2*b^5 + C*b^
7 - 2*(3*C*a^6*b + (A - 5*C)*a^4*b^3 - (A - 2*C)*a^2*b^5)*cos(d*x + c)^2 - 3*(C*a^5*b^2 - 2*C*a^3*b^4 + C*a*b^
6)*cos(d*x + c))*sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*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 ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39989, size = 483, normalized size = 1.78 \begin{align*} -\frac{\frac{4 \,{\left (3 \, C a^{5} + A a^{3} b^{2} - 4 \, C a^{3} b^{2} - 2 \, A a b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{4 \,{\left (C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}} - \frac{{\left (6 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac{{\left (6 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac{2 \,{\left (4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(3*C*a^5 + A*a^3*b^2 - 4*C*a^3*b^2 - 2*A*a*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + ar
ctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^2))
- 4*(C*a^4*tan(1/2*d*x + 1/2*c) + A*a^2*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 -
 b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (6*C*a^2 + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*
C*a^2 + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*C*a*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/
2*d*x + 1/2*c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3)
)/d