3.599 \(\int \frac {(1-\cos ^2(c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=117 \[ \frac {2 b \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d}-\frac {b \tan (c+d x)}{a^2 d}-\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3056, 3055, 3001, 3770, 2659, 205} \[ -\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {2 b \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d}-\frac {b \tan (c+d x)}{a^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x]),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [B]  time = 1.10, size = 236, normalized size = 2.02 \[ \frac {8 b \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 a b \tan (c+d x)-4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.52, size = 373, normalized size = 3.19 \[ \left [\frac {2 \, \sqrt {-a^{2} + b^{2}} b \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a b \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )}{4 \, a^{3} d \cos \left (d x + c\right )^{2}}, \frac {4 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a b \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )}{4 \, a^{3} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.86, size = 219, normalized size = 1.87 \[ -\frac {\frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

-1/2*((a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^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)))/(sqrt(a^2 - b^2)*a^3) - 2*(a*tan(1/2*d*x + 1/2*c)^3 + 2*b*tan(1/2*d*x
+ 1/2*c)^3 + a*tan(1/2*d*x + 1/2*c) - 2*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2))/d

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maple [B]  time = 0.17, size = 309, normalized size = 2.64 \[ \frac {2 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {1}{2 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d a}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}}{d \,a^{3}}-\frac {1}{2 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*b/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/d*b^3/a^3/((a-b)*(a+b))^(1/
2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+1/2/d/a/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/a/(tan(1/2*d*x+
1/2*c)-1)+1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*b+1/2/d/a*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*b
^2-1/2/d/a/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/a/(tan(1/2*d*x+1/2*c)+1)+1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*b-1/2/d/a*ln
(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*b^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((1-cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),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 = 1.74, size = 176, normalized size = 1.50 \[ \frac {\sin \left (c+d\,x\right )}{2\,a\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d}-\frac {b\,\sin \left (c+d\,x\right )}{a^2\,d\,\cos \left (c+d\,x\right )}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(c + d*x)/(2*a*d*cos(c + d*x)^2) - atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) + (2*b^2*atanh(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^3*d) - (b*sin(c + d*x))/(a^2*d*cos(c + d*x)) - (2*b*atanh((sin(c/2 + (d*x
)/2)*(b^2 - a^2)^(1/2))/(a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c)),x)

[Out]

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

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