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

Optimal. Leaf size=180 \[ -\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 A b+a^3 B-4 a b^2 B+2 A b^3\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{a (A b-a B) \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.335514, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4009, 4003, 12, 3831, 2659, 208} \[ -\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 A b+a^3 B-4 a b^2 B+2 A b^3\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{a (A b-a B) \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4009

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

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) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx &=\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\int \frac{\sec (c+d x) \left (-2 b (A b-a B)+\left (a A b+a^2 B-2 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{b \left (3 a A b-a^2 B-2 b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (3 a A b-a^2 B-2 b^2 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 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\left (3 a A b-a^2 B-2 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)^{5/2} (a+b)^{5/2} d}+\frac{a (A b-a B) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.677794, size = 157, normalized size = 0.87 \[ \frac{\frac{\left (2 a^2 A-3 a b B+A b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}-\frac{2 \left (a^2 B-3 a A b+2 b^2 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{(a B-A b) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.084, size = 238, normalized size = 1.3 \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 ) ^{2}} \left ( -1/2\,{\frac{ \left ( 2\,{a}^{2}A+Aab+2\,A{b}^{2}-B{a}^{2}-4\,Bab \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+1/2\,{\frac{ \left ( 2\,{a}^{2}A-Aab+2\,A{b}^{2}+B{a}^{2}-4\,Bab \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }} \right ) }-{\frac{3\,Aab-B{a}^{2}-2\,B{b}^{2}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}{\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+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x)

[Out]

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

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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+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 = 0.656479, size = 1631, normalized size = 9.06 \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*((B*a^2*b^2 - 3*A*a*b^3 + 2*B*b^4 + (B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2)*cos(d*x + c)^2 + 2*(B*a^3*b - 3*A*a
^2*b^2 + 2*B*a*b^3)*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*s
qrt(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^5 + A*a^4*b - 5*B*a^3*b^2 + A*a^2*b^3 + 4*B*a*b^4 - 2*A*b^5 + (2*A*a^5 - 3*B*a^4*b - A*a^3*b^2 + 3
*B*a^2*b^3 - A*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 +
2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d), 1/2*((B
*a^2*b^2 - 3*A*a*b^3 + 2*B*b^4 + (B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2)*cos(d*x + c)^2 + 2*(B*a^3*b - 3*A*a^2*b^2 +
2*B*a*b^3)*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))) + (B*a^5 + A*a^4*b - 5*B*a^3*b^2 + A*a^2*b^3 + 4*B*a*b^4 - 2*A*b^5 + (2*A*a^5 - 3*B*a^4*b - A*a^3*b^2 +
3*B*a^2*b^3 - A*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2
+ 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*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 ^{2}{\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)**2*(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)**2/(a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.44321, size = 540, normalized size = 3. \begin{align*} \frac{\frac{{\left (B a^{2} - 3 \, A a b + 2 \, B b^{2}\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} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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