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

Optimal. Leaf size=192 \[ -\frac {b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

[Out]

-b*(4*a^2+b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d+1/3*a*tan(d*x+c)/
(a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*(2*a^2+3*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*a*(2*a^2+13*b
^2)*tan(d*x+c)/(a^2-b^2)^3/d/(a+b*sec(d*x+c))

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Rubi [A]  time = 0.31, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3836, 4003, 12, 3831, 2659, 208} \[ -\frac {b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(4*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d)) + (a*
Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((2*a^2 + 3*b^2)*Tan[c + d*x])/(6*(a^2 - b^2)^2*d*(a
+ b*Sec[c + d*x])^2) + (a*(2*a^2 + 13*b^2)*Tan[c + d*x])/(6*(a^2 - b^2)^3*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 3836

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^4} \, dx &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) (-3 b+2 a \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (10 a b-\left (2 a^2+3 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\int -\frac {3 b \left (4 a^2+b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (b \left (4 a^2+b^2\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2+b^2\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2+b^2\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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 164, normalized size = 0.85 \[ \frac {\frac {6 b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {\sin (c+d x) \left (2 a^3 b^2+a \left (6 a^4+10 a^2 b^2-b^4\right ) \cos ^2(c+d x)-3 b \left (-2 a^4-9 a^2 b^2+b^4\right ) \cos (c+d x)+13 a b^4\right )}{(a-b)^3 (a+b)^3 (a \cos (c+d x)+b)^3}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*b*(4*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + ((2*a^3*b^2 + 13
*a*b^4 - 3*b*(-2*a^4 - 9*a^2*b^2 + b^4)*Cos[c + d*x] + a*(6*a^4 + 10*a^2*b^2 - b^4)*Cos[c + d*x]^2)*Sin[c + d*
x])/((a - b)^3*(a + b)^3*(b + a*Cos[c + d*x])^3))/(6*d)

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fricas [B]  time = 0.58, size = 901, normalized size = 4.69 \[ \left [-\frac {3 \, {\left (4 \, a^{2} b^{4} + b^{6} + {\left (4 \, a^{5} b + a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\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 (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6} + {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}, -\frac {3 \, {\left (4 \, a^{2} b^{4} + b^{6} + {\left (4 \, a^{5} b + a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\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 (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6} + {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[-1/12*(3*(4*a^2*b^4 + b^6 + (4*a^5*b + a^3*b^3)*cos(d*x + c)^3 + 3*(4*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + 3*(
4*a^3*b^3 + a*b^5)*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*sq
rt(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*(2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6 + (6*a^7 + 4*a^5*b^2 - 11*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + 3*(2*a^6*
b + 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3
*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^
2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9
+ b^11)*d), -1/6*(3*(4*a^2*b^4 + b^6 + (4*a^5*b + a^3*b^3)*cos(d*x + c)^3 + 3*(4*a^4*b^2 + a^2*b^4)*cos(d*x +
c)^2 + 3*(4*a^3*b^3 + a*b^5)*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))) - (2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6 + (6*a^7 + 4*a^5*b^2 - 11*a^3*b^4 + a*b^6)*cos(d*
x + c)^2 + 3*(2*a^6*b + 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b
^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*
x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*
a^4*b^7 - 4*a^2*b^9 + b^11)*d)]

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giac [B]  time = 0.35, size = 431, normalized size = 2.24 \[ -\frac {\frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(4*a^2*b + 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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) + (6*a^5*tan
(1/2*d*x + 1/2*c)^5 - 6*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 27*a^2*b^3*tan(1/2*
d*x + 1/2*c)^5 + 12*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*a^5*tan(1/2*d*x + 1/2*c)^
3 - 16*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 28*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*a^5*tan(1/2*d*x + 1/2*c) + 6*a^4*b
*tan(1/2*d*x + 1/2*c) + 12*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*a*b^4*tan(1/2*d
*x + 1/2*c) - 3*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*t
an(1/2*d*x + 1/2*c)^2 - a - b)^3))/d

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maple [A]  time = 0.43, size = 297, normalized size = 1.55 \[ \frac {\frac {-\frac {\left (2 a^{3}+2 a^{2} b +6 b^{2} a +b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {4 \left (3 a^{2}+7 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3}-2 a^{2} b +6 b^{2} a -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{3}}-\frac {b \left (4 a^{2}+b^{2}\right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sec(d*x+c))^4,x)

[Out]

1/d*(2*(-1/2*(2*a^3+2*a^2*b+6*a*b^2+b^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5+2/3*(3*a^2+7*b^2
)*a/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(2*a^3-2*a^2*b+6*a*b^2-b^3)/(a+b)/(a^3-3*a^2*b+3*
a*b^2-b^3)*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)^3-b*(4*a^2+b^2)/(a^6-3*a^4*
b^2+3*a^2*b^4-b^6)/((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(sec(d*x+c)^2/(a+b*sec(d*x+c))^4,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 = 4.34, size = 382, normalized size = 1.99 \[ \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^3+2\,a^2\,b+6\,a\,b^2+b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^3+7\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3-2\,a^2\,b+6\,a\,b^2-b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atanh}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (4\,a^2\,b+b^3\right )}\right )\,\left (4\,a^2+b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + b/cos(c + d*x))^4),x)

[Out]

((tan(c/2 + (d*x)/2)^5*(6*a*b^2 + 2*a^2*b + 2*a^3 + b^3))/((a + b)^3*(a - b)) - (4*tan(c/2 + (d*x)/2)^3*(7*a*b
^2 + 3*a^3))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)*(6*a*b^2 - 2*a^2*b + 2*a^3 - b^3))/((a +
b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(c/2 +
(d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - tan(c/2 + (d*x)/2)^6*(3*a*b^2
- 3*a^2*b + a^3 - b^3))) - (b*atanh((b*tan(c/2 + (d*x)/2)*(4*a^2 + b^2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3
- b^3))/(2*(a + b)^(1/2)*(a - b)^(7/2)*(4*a^2*b + b^3)))*(4*a^2 + b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*sec(d*x+c))**4,x)

[Out]

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

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