3.505 \(\int \frac{\cos ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=261 \[ \frac{\left (7 a^2 b^2+2 a^4-12 b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}+\frac{\left (a^2-4 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{b \left (a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{b x \left (a^2+4 b^2\right )}{a^5} \]

[Out]

-((b*(a^2 + 4*b^2)*x)/a^5) + (2*b^4*(5*a^2 - 4*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*
(a - b)^(3/2)*(a + b)^(3/2)*d) + ((2*a^4 + 7*a^2*b^2 - 12*b^4)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*d) - (b*(a^2 -
2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 4*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*(a
^2 - b^2)*d) + (b^2*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.835403, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3847, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (7 a^2 b^2+2 a^4-12 b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}+\frac{\left (a^2-4 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{b \left (a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{b x \left (a^2+4 b^2\right )}{a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(a^2 + 4*b^2)*x)/a^5) + (2*b^4*(5*a^2 - 4*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*
(a - b)^(3/2)*(a + b)^(3/2)*d) + ((2*a^4 + 7*a^2*b^2 - 12*b^4)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*d) - (b*(a^2 -
2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 4*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*(a
^2 - b^2)*d) + (b^2*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 3847

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(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)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 4104

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 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 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{\cos ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^3(c+d x) \left (-a^2+4 b^2+a b \sec (c+d x)-3 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos ^2(c+d x) \left (-6 b \left (a^2-2 b^2\right )+a \left (2 a^2+b^2\right ) \sec (c+d x)+2 b \left (a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 \left (2 a^4+7 a^2 b^2-12 b^4\right )+2 a b \left (a^2+2 b^2\right ) \sec (c+d x)+6 b^2 \left (a^2-2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-6 b \left (a^4+3 a^2 b^2-4 b^4\right )-6 a b^2 \left (a^2-2 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{b \left (a^2+4 b^2\right ) x}{a^5}+\frac{\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b^4 \left (5 a^2-4 b^2\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{b \left (a^2+4 b^2\right ) x}{a^5}+\frac{\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b^3 \left (5 a^2-4 b^2\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{b \left (a^2+4 b^2\right ) x}{a^5}+\frac{\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 b^3 \left (5 a^2-4 b^2\right )\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 )}{a^5 \left (a^2-b^2\right ) d}\\ &=-\frac{b \left (a^2+4 b^2\right ) x}{a^5}+\frac{2 b^4 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (2 a^4+7 a^2 b^2-12 b^4\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b \left (a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-4 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.05719, size = 176, normalized size = 0.67 \[ \frac{9 a \left (a^2+4 b^2\right ) \sin (c+d x)+\frac{24 b^4 \left (4 b^2-5 a^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-6 a^2 b \sin (2 (c+d x))+a^3 \sin (3 (c+d x))+\frac{12 a b^5 \sin (c+d x)}{(b-a) (a+b) (a \cos (c+d x)+b)}-12 b (2 b-i a) (2 b+i a) (c+d x)}{12 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*b*((-I)*a + 2*b)*(I*a + 2*b)*(c + d*x) + (24*b^4*(-5*a^2 + 4*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqr
t[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 9*a*(a^2 + 4*b^2)*Sin[c + d*x] + (12*a*b^5*Sin[c + d*x])/((-a + b)*(a + b)*
(b + a*Cos[c + d*x])) - 6*a^2*b*Sin[2*(c + d*x)] + a^3*Sin[3*(c + d*x)])/(12*a^5*d)

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Maple [B]  time = 0.087, size = 508, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
5*b+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*b^2+4/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)^3+12/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*b^2+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*ta
n(1/2*d*x+1/2*c)-2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*b+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*ta
n(1/2*d*x+1/2*c)*b^2-2/d/a^3*b*arctan(tan(1/2*d*x+1/2*c))-8/d/a^5*arctan(tan(1/2*d*x+1/2*c))*b^3+2/d*b^5/a^4/(
a^2-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)+10/d*b^4/a^3/(a+b)/(a-b)/((a+b
)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-8/d*b^6/a^5/(a+b)/(a-b)/((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(cos(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.23767, size = 1661, normalized size = 6.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(6*(a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*d*x*cos(d*x + c) + 6*(a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b
^8)*d*x - 3*(5*a^2*b^5 - 4*b^7 + (5*a^3*b^4 - 4*a*b^6)*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*(2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7 + (a^8 - 2*a^6*b^2 + a^
4*b^4)*cos(d*x + c)^3 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + 2*(a^8 + a^6*b^2 - 5*a^4*b^4 + 3*a^2*
b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b^3 + a^5*b^5)*
d), -1/3*(3*(a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*d*x*cos(d*x + c) + 3*(a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 +
4*b^8)*d*x - 3*(5*a^2*b^5 - 4*b^7 + (5*a^3*b^4 - 4*a*b^6)*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^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7 + (a^8 - 2
*a^6*b^2 + a^4*b^4)*cos(d*x + c)^3 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)^2 + 2*(a^8 + a^6*b^2 - 5*a^4
*b^4 + 3*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b^
3 + a^5*b^5)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.32735, size = 452, normalized size = 1.73 \begin{align*} \frac{\frac{6 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - a^{4} b^{2}\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 )}} + \frac{6 \,{\left (5 \, a^{2} b^{4} - 4 \, b^{6}\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^{7} - a^{5} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{3 \,{\left (a^{2} b + 4 \, b^{3}\right )}{\left (d x + c\right )}}{a^{5}} + \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*b^5*tan(1/2*d*x + 1/2*c)/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)
) + 6*(5*a^2*b^4 - 4*b^6)*(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^7 - a^5*b^2)*sqrt(-a^2 + b^2)) - 3*(a^2*b + 4*b^3)*(d*x + c)/
a^5 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d*x + 1/2*c)^5 + 2*a^2*ta
n(1/2*d*x + 1/2*c)^3 + 18*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c)
+ 9*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^4))/d