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

Optimal. Leaf size=130 \[ -\frac {6 a b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )^2} \]

[Out]

-6*a*b^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)/d+b*sec(d*x+c)/(a^2-b^2)/d/(a+b*sin(
d*x+c))-sec(d*x+c)*(3*a*b-(a^2+2*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.21, antiderivative size = 130, 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 = {2694, 2866, 12, 2660, 618, 204} \[ -\frac {6 a b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) + (b*Sec[c + d*x])/((a^2 - b
^2)*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]*(3*a*b - (a^2 + 2*b^2)*Sin[c + d*x]))/((a^2 - b^2)^2*d)

Rule 12

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2694

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m +
1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; F
reeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {\sec ^2(c+d x) (-a+2 b \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{-a^2+b^2}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac {\int -\frac {3 a b^2}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}-\frac {\left (3 a b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}-\frac {\left (6 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac {\left (12 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {6 a b^2 \tan ^{-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} d}+\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 a b-\left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.15, size = 162, normalized size = 1.25 \[ \frac {-\frac {6 a b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {b^3 \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + Sin[(c + d*x)/2]*(1/((a + b)^
2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) - (b^3*Cos[c +
d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])))/d

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fricas [A]  time = 0.50, size = 538, normalized size = 4.14 \[ \left [-\frac {2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*a^4*b - 4*a^2*b^3 + 2*b^5 + 2*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + 3*(a*b^3*cos(d*x + c)*sin(d*
x + c) + a^2*b^2*cos(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2
- b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x
+ c) - a^2 - b^2)) - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d
*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)), -(a^4*b - 2*a^2*b^3 + b^5 + (a^4
*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^2 - 3*(a*b^3*cos(d*x + c)*sin(d*x + c) + a^2*b^2*cos(d*x + c))*sqrt(a^2 - b
^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))/((a
^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(
d*x + c))]

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giac [B]  time = 2.55, size = 271, normalized size = 2.08 \[ -\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} b - a b^{3}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a*b^2/(
(a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (a^4*tan(1/2*d*x + 1/2*c)^3 + a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + b^4*
tan(1/2*d*x + 1/2*c)^3 + 3*a*b^3*tan(1/2*d*x + 1/2*c)^2 + a^4*tan(1/2*d*x + 1/2*c) - 3*a^2*b^2*tan(1/2*d*x + 1
/2*c) - b^4*tan(1/2*d*x + 1/2*c) - 2*a^3*b - a*b^3)/((a^5 - 2*a^3*b^2 + a*b^4)*(a*tan(1/2*d*x + 1/2*c)^4 + 2*b
*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c) - a)))/d

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maple [A]  time = 0.22, size = 222, normalized size = 1.71 \[ -\frac {1}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a}-\frac {2 b^{3}}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {6 b^{2} a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*sin(d*x+c))^2,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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.40, size = 303, normalized size = 2.33 \[ -\frac {\frac {2\,\left (2\,a^2\,b+b^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {6\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-a^4+3\,a^2\,b^2+b^4\right )}{a\,{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^4+a^2\,b^2+b^4\right )}{a\,{\left (a^2-b^2\right )}^2}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {6\,a\,b^2\,\mathrm {atan}\left (\frac {\frac {3\,a\,b^2\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {6\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{6\,a\,b^2}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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