3.330 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=220 \[ \frac{\left (a^2 A b^3+5 a^3 b^2 B-2 a^5 B-6 a b^4 B+2 A b^5\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-3 a^3 B+6 a b^2 B-4 A b^3\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{B \tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

[Out]

(B*ArcTanh[Sin[c + d*x]])/(b^3*d) + ((a^2*A*b^3 + 2*A*b^5 - 2*a^5*B + 5*a^3*b^2*B - 6*a*b^4*B)*ArcTanh[(Sqrt[a
- b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*(A*b - a*B)*Tan[c + d*x])/(2*
b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (a*(a^2*A*b - 4*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Tan[c + d*x])/(2*b^2*
(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.686463, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4028, 4080, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (a^2 A b^3+5 a^3 b^2 B-2 a^5 B-6 a b^4 B+2 A b^5\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-3 a^3 B+6 a b^2 B-4 A b^3\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{B \tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(b^3*d) + ((a^2*A*b^3 + 2*A*b^5 - 2*a^5*B + 5*a^3*b^2*B - 6*a*b^4*B)*ArcTanh[(Sqrt[a
- b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*(A*b - a*B)*Tan[c + d*x])/(2*
b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (a*(a^2*A*b - 4*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Tan[c + d*x])/(2*b^2*
(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4028

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(a^2*(A*b - a*B)*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[a*b*(A*b - a*B)*(m
+ 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Csc[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && 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 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) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx &=-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (-2 a b (A b-a B)-\left (a^2-2 b^2\right ) (A b-a B) \sec (c+d x)-2 b \left (a^2-b^2\right ) B \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \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 (b^2 \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right )+2 b \left (a^2-b^2\right )^2 B \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{B \int \sec (c+d x) \, dx}{b^3}+\frac{\left (a^2 A b^3+2 A b^5-2 a^5 B+5 a^3 b^2 B-6 a b^4 B\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{B \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (a^2 A b^3+2 A b^5-2 a^5 B+5 a^3 b^2 B-6 a b^4 B\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{B \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (a^2 A b^3+2 A b^5-2 a^5 B+5 a^3 b^2 B-6 a b^4 B\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^4 \left (a^2-b^2\right )^2 d}\\ &=\frac{B \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{\left (a^2 A b^3+2 A b^5-2 a^5 B+5 a^3 b^2 B-6 a b^4 B\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^3 (a+b)^{5/2} d}-\frac{a^2 (A b-a B) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \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 = 1.82697, size = 270, normalized size = 1.23 \[ \frac{\cos (c+d x) (A+B \sec (c+d x)) \left (\frac{a b \left (-2 a^3 B+5 a b^2 B-3 A b^3\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{2 \left (-a^2 A b^3-5 a^3 b^2 B+2 a^5 B+6 a b^4 B-2 A b^5\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^2 (a B-A b) \sin (c+d x)}{(b-a) (a+b) (a \cos (c+d x)+b)^2}-2 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 b^3 d (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.095, size = 1085, normalized size = 4.9 \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 43.5527, size = 3051, normalized size = 13.87 \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*((2*B*a^5*b^2 - 5*B*a^3*b^4 - A*a^2*b^5 + 6*B*a*b^6 - 2*A*b^7 + (2*B*a^7 - 5*B*a^5*b^2 - A*a^4*b^3 + 6*B
*a^3*b^4 - 2*A*a^2*b^5)*cos(d*x + c)^2 + 2*(2*B*a^6*b - 5*B*a^4*b^3 - A*a^3*b^4 + 6*B*a^2*b^5 - 2*A*a*b^6)*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*(B*a^6*b^2 - 3*B
*a^4*b^4 + 3*B*a^2*b^6 - B*b^8 + (B*a^8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*cos(d*x + c)^2 + 2*(B*a^7*b -
3*B*a^5*b^3 + 3*B*a^3*b^5 - B*a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a
^2*b^6 - B*b^8 + (B*a^8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*cos(d*x + c)^2 + 2*(B*a^7*b - 3*B*a^5*b^3 + 3
*B*a^3*b^5 - B*a*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(3*B*a^6*b^2 - A*a^5*b^3 - 9*B*a^4*b^4 + 5*A*a^
3*b^5 + 6*B*a^2*b^6 - 4*A*a*b^7 + (2*B*a^7*b - 7*B*a^5*b^3 + 3*A*a^4*b^4 + 5*B*a^3*b^5 - 3*A*a^2*b^6)*cos(d*x
+ c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3
*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d), -1/2*((2*B*a^5*b^2 - 5*B*a^3*
b^4 - A*a^2*b^5 + 6*B*a*b^6 - 2*A*b^7 + (2*B*a^7 - 5*B*a^5*b^2 - A*a^4*b^3 + 6*B*a^3*b^4 - 2*A*a^2*b^5)*cos(d*
x + c)^2 + 2*(2*B*a^6*b - 5*B*a^4*b^3 - A*a^3*b^4 + 6*B*a^2*b^5 - 2*A*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*ar
ctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b
^6 - B*b^8 + (B*a^8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*cos(d*x + c)^2 + 2*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a
^3*b^5 - B*a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + (B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b^6 - B*b^8 + (B*a^
8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*cos(d*x + c)^2 + 2*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a^3*b^5 - B*a*b^7)*
cos(d*x + c))*log(-sin(d*x + c) + 1) + (3*B*a^6*b^2 - A*a^5*b^3 - 9*B*a^4*b^4 + 5*A*a^3*b^5 + 6*B*a^2*b^6 - 4*
A*a*b^7 + (2*B*a^7*b - 7*B*a^5*b^3 + 3*A*a^4*b^4 + 5*B*a^3*b^5 - 3*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^
8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos
(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*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 + B \sec{\left (c + d x \right )}\right ) \sec ^{3}{\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)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 1.51929, size = 656, normalized size = 2.98 \begin{align*} -\frac{\frac{{\left (2 \, B a^{5} - 5 \, B a^{3} b^{2} - A a^{2} b^{3} + 6 \, B a b^{4} - 2 \, A b^{5}\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^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac{2 \, B a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\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 )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*B*a^5 - 5*B*a^3*b^2 - A*a^2*b^3 + 6*B*a*b^4 - 2*A*b^5)*(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^3 - 2*a^2*b^5 + b^7)*sq
rt(-a^2 + b^2)) - B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - (2*B*a
^5*tan(1/2*d*x + 1/2*c)^3 - 3*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 + A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*B*a^3*b^2*
tan(1/2*d*x + 1/2*c)^3 + 3*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 4*A*a*b^4*t
an(1/2*d*x + 1/2*c)^3 - 2*B*a^5*tan(1/2*d*x + 1/2*c) - 3*B*a^4*b*tan(1/2*d*x + 1/2*c) + A*a^3*b^2*tan(1/2*d*x
+ 1/2*c) + 5*B*a^3*b^2*tan(1/2*d*x + 1/2*c) - 3*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*B*a^2*b^3*tan(1/2*d*x + 1/2
*c) - 4*A*a*b^4*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*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)^2))/d