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

Optimal. Leaf size=164 \[ \frac {\left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

[Out]

(2*B*a^2+B*b^2-3*C*a*b)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*(B*b
-C*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1/2*(3*B*a*b-C*a^2-2*C*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(
d*x+c))

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Rubi [A]  time = 0.26, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4060, 12, 3831, 2659, 208} \[ \frac {\left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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 4060

Int[((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))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 1.14, size = 172, normalized size = 1.05 \[ \frac {-\frac {2 \left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\left (2 a^3 C-4 a^2 b B+a b^2 C+b^3 B\right ) \sin (c+d x)}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac {b (b B-a C) \sin (c+d x)}{a (a-b) (a+b) (a \cos (c+d x)+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.98, size = 752, normalized size = 4.59 \[ \left [\frac {{\left (2 \, B a^{2} b^{2} - 3 \, C a b^{3} + B b^{4} + {\left (2 \, B a^{4} - 3 \, C a^{3} b + B a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, B a^{3} b - 3 \, C a^{2} b^{2} + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (C a^{4} b - 3 \, B a^{3} b^{2} + C a^{2} b^{3} + 3 \, B a b^{4} - 2 \, C b^{5} + {\left (2 \, C a^{5} - 4 \, B a^{4} b - C a^{3} b^{2} + 5 \, B a^{2} b^{3} - C a b^{4} - B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}, \frac {{\left (2 \, B a^{2} b^{2} - 3 \, C a b^{3} + B b^{4} + {\left (2 \, B a^{4} - 3 \, C a^{3} b + B a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, B a^{3} b - 3 \, C a^{2} b^{2} + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{4} b - 3 \, B a^{3} b^{2} + C a^{2} b^{3} + 3 \, B a b^{4} - 2 \, C b^{5} + {\left (2 \, C a^{5} - 4 \, B a^{4} b - C a^{3} b^{2} + 5 \, B a^{2} b^{3} - C a b^{4} - B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((2*B*a^2*b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C
*a^2*b^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*(C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2*b
^3 - C*a*b^4 - B*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
*B*a^2*b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C*a^2*b^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))) + (C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2
*b^3 - C*a*b^4 - B*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)]

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giac [B]  time = 0.36, size = 399, normalized size = 2.43 \[ \frac {\frac {{\left (2 \, B a^{2} - 3 \, C a b + B b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.91, size = 236, normalized size = 1.44 \[ \frac {-\frac {2 \left (-\frac {\left (4 B a b +b^{2} B -2 a^{2} C -C a b -2 b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 B a b -b^{2} B -2 a^{2} C +C a b -2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{2}}+\frac {\left (2 a^{2} B +b^{2} B -3 C a b \right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 7.26, size = 251, normalized size = 1.53 \[ \frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,B\,a^2-3\,C\,a\,b+B\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,C\,a^2-B\,b^2+2\,C\,b^2-4\,B\,a\,b+C\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,b^2+2\,C\,a^2+2\,C\,b^2-4\,B\,a\,b-C\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + b/cos(c + d*x))^3,x)

[Out]

(atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1/2)*(a - b)^(5/2)))*(2*B*a^2 + B*b^2
- 3*C*a*b))/(d*(a + b)^(5/2)*(a - b)^(5/2)) - ((tan(c/2 + (d*x)/2)^3*(2*C*a^2 - B*b^2 + 2*C*b^2 - 4*B*a*b + C*
a*b))/((a + b)^2*(a - b)) - (tan(c/2 + (d*x)/2)*(B*b^2 + 2*C*a^2 + 2*C*b^2 - 4*B*a*b - C*a*b))/((a + b)*(a^2 -
2*a*b + b^2)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) +
a^2 + b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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