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

Optimal. Leaf size=177 \[ \frac{\left (a^2 (2 A+C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a \left (a^2 (-C)+3 A b^2+4 b^2 C\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

[Out]

((a^2*(2*A + C) + b^2*(A + 2*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(
5/2)*d) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - (a*(3*A*b^2 - a^2*C + 4*
b^2*C)*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 = 0.321387, antiderivative size = 177, 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 = {4081, 4003, 12, 3831, 2659, 208} \[ \frac{\left (a^2 (2 A+C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a \left (a^2 (-C)+3 A b^2+4 b^2 C\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4081

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(
m_), x_Symbol] :> -Simp[((A*b^2 + 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[a*b*(A + C)*(m + 1) -
(A*b^2 + a^2*C + b*(A*b + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 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 (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \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 a b (A+C)+\left (A b^2-a^2 C+2 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (3 A b^2-a^2 C+4 b^2 C\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 (a^2 (2 A+C)+b^2 (A+2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (3 A b^2-a^2 C+4 b^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (a^2 (2 A+C)+b^2 (A+2 C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (3 A b^2-a^2 C+4 b^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (a^2 (2 A+C)+b^2 (A+2 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (3 A b^2-a^2 C+4 b^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (a^2 (2 A+C)+b^2 (A+2 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 )}{b \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (2 a^2 A+A b^2+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)^{5/2} (a+b)^{5/2} d}-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (3 A b^2-a^2 C+4 b^2 C\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.39468, size = 342, normalized size = 1.93 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{a \sec (c) \left (\left (a^2 b^2 (5 A+2 C)+a^4 C-2 A b^4\right ) \sin (2 c+d x)+a b \left (A b^2-a^2 (4 A+3 C)\right ) \sin (c+2 d x)+\sin (d x) \left (-a^2 b^2 (11 A+10 C)+a^4 C+2 A b^4\right )\right )+b \left (a^2+2 b^2\right ) \tan (c) \left (a^2 (4 A+3 C)-A b^2\right )}{\left (a^3-a b^2\right )^2}-\frac{4 i (\cos (c)-i \sin (c)) \left (a^2 (2 A+C)+b^2 (A+2 C)\right ) (a \cos (c+d x)+b)^2 \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{5/2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{2 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(((-4*I)*(a^2*(2*A + C) + b^2*(A + 2*C))*ArcTan[((I*
Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(b
+ a*Cos[c + d*x])^2*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^(5/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (a*Sec[c]*((2*A*b^4
+ a^4*C - a^2*b^2*(11*A + 10*C))*Sin[d*x] + (-2*A*b^4 + a^4*C + a^2*b^2*(5*A + 2*C))*Sin[2*c + d*x] + a*b*(A*
b^2 - a^2*(4*A + 3*C))*Sin[c + 2*d*x]) + b*(a^2 + 2*b^2)*(-(A*b^2) + a^2*(4*A + 3*C))*Tan[c])/(a^3 - a*b^2)^2)
)/(2*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [A]  time = 0.093, size = 230, 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 ( 4\,Aab+A{b}^{2}+{a}^{2}C+4\,abC \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 ( 4\,Aab-A{b}^{2}-{a}^{2}C+4\,abC \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }} \right ) }+{\frac{2\,{a}^{2}A+A{b}^{2}+{a}^{2}C+2\,{b}^{2}C}{{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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

1/d*(-2*(-1/2*(4*A*a*b+A*b^2+C*a^2+4*C*a*b)/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(4*A*a*b-A*b^2-C*a^
2+4*C*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+(2*
A*a^2+A*b^2+C*a^2+2*C*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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.647531, size = 1577, normalized size = 8.91 \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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((2*A + C)*a^2*b^2 + (A + 2*C)*b^4 + ((2*A + C)*a^4 + (A + 2*C)*a^2*b^2)*cos(d*x + c)^2 + 2*((2*A + C)*a
^3*b + (A + 2*C)*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*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*(C*a^5 - (3*A + 5*C)*a^3*b^2 + (3*A + 4*C)*a*b^4 - ((4*A + 3*C)*a^4*b - (5*A + 3*C)*a^2*b^3 + A*b^5
)*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*(((2*A + C)*a^2*b^2 + (
A + 2*C)*b^4 + ((2*A + C)*a^4 + (A + 2*C)*a^2*b^2)*cos(d*x + c)^2 + 2*((2*A + C)*a^3*b + (A + 2*C)*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))) + (C*a^5
- (3*A + 5*C)*a^3*b^2 + (3*A + 4*C)*a*b^4 - ((4*A + 3*C)*a^4*b - (5*A + 3*C)*a^2*b^3 + A*b^5)*cos(d*x + c))*si
n(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 + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\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)*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 1.28683, size = 501, normalized size = 2.83 \begin{align*} \frac{\frac{{\left (2 \, A a^{2} + C a^{2} + A b^{2} + 2 \, C 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{C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C 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 ) - 4 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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