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

Optimal. Leaf size=272 \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{\left (-6 a^2 B+4 a A b-b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b^2 d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.86559, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {4029, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{\left (-6 a^2 B+4 a A b-b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b^2 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4029

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

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) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac{a (A b-a B) \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 a (A b-a B)-b (A b-a B) \sec (c+d x)-\left (2 a A b-3 a^2 B+b^2 B\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 A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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 A b-3 a^2 B+b^2 B\right )+b \left (2 a A b-a^2 B-b^2 B\right ) \sec (c+d x)+2 \left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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 (2 a A b-3 a^2 B+b^2 B\right )-\left (a^2-b^2\right ) \left (4 a A b-6 a^2 B-b^2 B\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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 (4 a A b-6 a^2 B-b^2 B\right ) \int \sec (c+d x) \, dx}{2 b^4}+\frac{\left (a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\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 (4 a A b-6 a^2 B-b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\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 (4 a A b-6 a^2 B-b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\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{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \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 = 6.27154, size = 438, normalized size = 1.61 \[ \frac{a^4 B \sin (c+d x)-a^3 A b \sin (c+d x)}{b^3 d (b-a) (a+b) (a \cos (c+d x)+b)}-\frac{2 a^2 \left (-2 a^2 A b+3 a^3 B-4 a b^2 B+3 A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 d \sqrt{a^2-b^2} \left (b^2-a^2\right )}+\frac{\left (-6 a^2 B+4 a A b-b^2 B\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac{\left (6 a^2 B-4 a A b+b^2 B\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac{A b \sin \left (\frac{1}{2} (c+d x)\right )-2 a B \sin \left (\frac{1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{A b \sin \left (\frac{1}{2} (c+d x)\right )-2 a B \sin \left (\frac{1}{2} (c+d x)\right )}{b^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{B}{4 b^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{B}{4 b^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.098, size = 698, normalized size = 2.6 \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)

[Out]

-2/d*a^3/b^2/(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)*B+4/d*a^4/b^3/(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-6/d*a^2/b/(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-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))*B+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))*B-1/2/d*B/b^2/(tan(1/2*d*x+1/2*c)+1)^2-1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*A+2
/d/b^3/(tan(1/2*d*x+1/2*c)+1)*B*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)*B-2/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*A*a+3/d/
b^4*ln(tan(1/2*d*x+1/2*c)+1)*B*a^2+1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*B+1/2/d*B/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1
/d/b^2/(tan(1/2*d*x+1/2*c)-1)*A+2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*B*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*B+2/d/b^3*
ln(tan(1/2*d*x+1/2*c)-1)*A*a-3/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)*B*a^2-1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*B

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 49.3933, size = 2969, normalized size = 10.92 \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(2*((3*B*a^6 - 2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3)*cos(d*x + c)^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3
*b^3 + 3*A*a^2*b^4)*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*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*cos(d*x + c)^
3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*cos(d*x + c)^2)*l
og(sin(d*x + c) + 1) - ((6*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^
6)*cos(d*x + c)^3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*c
os(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(B*a^4*b^3 - 2*B*a^2*b^5 + B*b^7 - 2*(3*B*a^6*b - 2*A*a^5*b^2 - 5*B*
a^4*b^3 + 3*A*a^3*b^4 + 2*B*a^2*b^5 - A*a*b^6)*cos(d*x + c)^2 - (3*B*a^5*b^2 - 2*A*a^4*b^3 - 6*B*a^3*b^4 + 4*A
*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*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*B*a^6 - 2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3)*cos(
d*x + c)^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*b^4)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*arctan(-sq
rt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - ((6*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^2 + 8*A*a
^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*cos(d*x + c)^3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a
^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((6*B*a^7 - 4*A*a^6*b - 11*B
*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*cos(d*x + c)^3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B
*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(B*a^4*b^
3 - 2*B*a^2*b^5 + B*b^7 - 2*(3*B*a^6*b - 2*A*a^5*b^2 - 5*B*a^4*b^3 + 3*A*a^3*b^4 + 2*B*a^2*b^5 - A*a*b^6)*cos(
d*x + c)^2 - (3*B*a^5*b^2 - 2*A*a^4*b^3 - 6*B*a^3*b^4 + 4*A*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*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)]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.25255, size = 518, normalized size = 1.9 \begin{align*} -\frac{\frac{4 \,{\left (3 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\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 (B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{3} b \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 \, B a^{2} - 4 \, A a b + B 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 \, B a^{2} - 4 \, A a b + B 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 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B 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)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*(3*B*a^5 - 2*A*a^4*b - 4*B*a^3*b^2 + 3*A*a^2*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^2*b^4 - b^6)*sqrt(-a^2 + b^2)
) - 4*(B*a^4*tan(1/2*d*x + 1/2*c) - A*a^3*b*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*B*a^2 - 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*
B*a^2 - 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*B*a*tan(1/2*d*x + 1/2*c)^3 - 2*A*b*tan(
1/2*d*x + 1/2*c)^3 + B*b*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*tan(1/2*d*x + 1/2*c) + 2*A*b*tan(1/2*d*x + 1/2*c) + B*
b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3))/d