3.6.9 \(\int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [509]

Optimal. Leaf size=134 \[ -\frac {3 a b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]

[Out]

-3*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*a*tan(d*x+c)/(a^2-b^2
)/d/(a+b*sec(d*x+c))^2+1/2*(a^2+2*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 = 0.14, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3921, 4088, 12, 3916, 2738, 214} \begin {gather*} \frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {a \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {3 a b \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) + (a*Tan[c + d*x]
)/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((a^2 + 2*b^2)*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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

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 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3921

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 4088

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] + D
ist[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))^3} \, dx &=\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) (-2 b+a \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {3 a b \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {(3 a b) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {(3 a) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {(3 a) \text {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 )^2 d}\\ &=-\frac {3 a b \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 \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2+2 b^2\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 = 0.41, size = 115, normalized size = 0.86 \begin {gather*} \frac {\frac {6 a b \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\left (b \left (a^2+2 b^2\right )+a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(b+a \cos (c+d x))^2}}{2 (a-b)^2 (a+b)^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 195, normalized size = 1.46

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (2 a^{2}+b a +2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 b a +b^{2}\right )}+\frac {\left (2 a^{2}-b a +2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 b a +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {3 b a \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(195\)
default	\(\frac {\frac {-\frac {\left (2 a^{2}+b a +2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 b a +b^{2}\right )}+\frac {\left (2 a^{2}-b a +2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 b a +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {3 b a \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(195\)
risch	\(\frac {i \left (3 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+4 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+2 a^{4}+b^{2} a^{2}\right )}{a \left (-a^{2}+b^{2}\right )^{2} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2}}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(316\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(2*(-1/2*(2*a^2+a*b+2*b^2)/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(2*a^2-a*b+2*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-b*tan(1/2*d*x+1/2*c)^2-a-b)^2-3*b*a/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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 de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (121) = 242\).
time = 3.22, size = 565, normalized size = 4.22 \begin {gather*} \left [\frac {3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} + 2 \, a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3}\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 (a^{4} b + a^{2} b^{3} - 2 \, b^{5} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\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 {3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} + 2 \, a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3}\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 (a^{4} b + a^{2} b^{3} - 2 \, b^{5} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(a^3*b*cos(d*x + c)^2 + 2*a^2*b^2*cos(d*x + c) + a*b^3)*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*(a^4*b + a^2*b^3 - 2*b^5 + (2*a^5 - a^3*b^2 - 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*(3*(a^3*b*cos(d*x + c)^2 + 2*a^2*b^2*cos(d*x
+ c) + a*b^3)*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))) - (a
^4*b + a^2*b^3 - 2*b^5 + (2*a^5 - a^3*b^2 - 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (121) = 242\).
time = 0.53, size = 277, normalized size = 2.07 \begin {gather*} \frac {\frac {3 \, {\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 )} a b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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*b/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - (2*a^3*tan(1/2*d*x + 1/2*c)^3 - a^2*b*tan
(1/2*d*x + 1/2*c)^3 + a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a^3*tan(1/2*d*x + 1/2*c)
- a^2*b*tan(1/2*d*x + 1/2*c) - a*b^2*tan(1/2*d*x + 1/2*c) - 2*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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Mupad [B]
time = 3.37, size = 210, normalized size = 1.57 \begin {gather*} -\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^2+a\,b+2\,b^2\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-a\,b+2\,b^2\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 )}-\frac {3\,a\,b\,\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 )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + b/cos(c + d*x))^3),x)

[Out]

- ((tan(c/2 + (d*x)/2)^3*(a*b + 2*a^2 + 2*b^2))/((a + b)^2*(a - b)) - (tan(c/2 + (d*x)/2)*(2*a^2 - a*b + 2*b^2
))/((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)) - (3*a*b*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))))/(d*(a + b)^(5/2)*(a - b)^(5/2))

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