3.609 \(\int \frac{(1-\cos ^2(c+d x)) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=195 \[ -\frac{2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}+\frac{b \left (a^2-4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{2 b \tan (c+d x) \sec (c+d x)}{a^3 d}+\frac{4 \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.916285, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}+\frac{b \left (a^2-4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{2 b \tan (c+d x) \sec (c+d x)}{a^3 d}+\frac{4 \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x])^2,x]

[Out]

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

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 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 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 3770

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (4 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-12 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)+8 b \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac{2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 \left (a^4-13 a^2 b^2+12 b^4\right )+4 a b \left (a^2-b^2\right ) \cos (c+d x)-12 b^2 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}-\frac{2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (6 b \left (a^4-5 a^2 b^2+4 b^4\right )-12 a b^2 \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}-\frac{2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\left (b \left (a^2-4 b^2\right )\right ) \int \sec (c+d x) \, dx}{a^5}-\frac{\left (b^2 \left (3 a^2-4 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^5}\\ &=\frac{b \left (a^2-4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}-\frac{2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac{\left (2 b^2 \left (3 a^2-4 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^5 d}\\ &=-\frac{2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 \sqrt{a-b} \sqrt{a+b} d}+\frac{b \left (a^2-4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}-\frac{2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac{4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.21069, size = 475, normalized size = 2.44 \[ \frac{b^3 \sin (c+d x)}{a^4 d (a+b \cos (c+d x))}+\frac{9 b^2 \sin \left (\frac{1}{2} (c+d x)\right )-a^2 \sin \left (\frac{1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{9 b^2 \sin \left (\frac{1}{2} (c+d x)\right )-a^2 \sin \left (\frac{1}{2} (c+d x)\right )}{3 a^4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 b^2 \left (3 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^5 d \sqrt{b^2-a^2}}+\frac{\left (4 b^3-a^2 b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{\left (a^2 b-4 b^3\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{a-6 b}{12 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{6 b-a}{12 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{6 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.069, size = 458, normalized size = 2.4 \begin{align*} 2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-6\,{\frac{{b}^{2}}{d{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+8\,{\frac{{b}^{4}}{d{a}^{5}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-3\,{\frac{{b}^{2}}{d{a}^{4} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{b}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+4\,{\frac{{b}^{3}\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{5}}}-{\frac{1}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{{b}^{2}}{d{a}^{4} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{b}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-4\,{\frac{{b}^{3}\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b^3/a^4*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)-6/d/a^3/((a+b)*(a-b))^(1/2)
*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^2+8/d*b^4/a^5/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2
*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/3/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2-1/d/a^3
/(tan(1/2*d*x+1/2*c)-1)^2*b-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*b-3/d*b^2/a^4/(tan(1/2*d*x+1/2*c)-1)-1/d*b/a^3*ln(t
an(1/2*d*x+1/2*c)-1)+4/d*b^3/a^5*ln(tan(1/2*d*x+1/2*c)-1)-1/3/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/a^2/(tan(1/
2*d*x+1/2*c)+1)^2+1/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2*b-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*b-3/d*b^2/a^4/(tan(1/2*d*x
+1/2*c)+1)+1/d*b/a^3*ln(tan(1/2*d*x+1/2*c)+1)-4/d*b^3/a^5*ln(tan(1/2*d*x+1/2*c)+1)

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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((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.78843, size = 1886, normalized size = 9.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/6*(3*((3*a^2*b^3 - 4*b^5)*cos(d*x + c)^4 + (3*a^3*b^2 - 4*a*b^4)*cos(d*x + c)^3)*sqrt(-a^2 + b^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)) + 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 + (a^
5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x +
c)^4 + (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) + 2*(a^6 - a^4*b^2 - (a^5*b - 13*a
^3*b^3 + 12*a*b^5)*cos(d*x + c)^3 - (a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c)^2 - 2*(a^5*b - a^3*b^3)*cos(d*x
+ c))*sin(d*x + c))/((a^7*b - a^5*b^3)*d*cos(d*x + c)^4 + (a^8 - a^6*b^2)*d*cos(d*x + c)^3), -1/6*(6*((3*a^2*
b^3 - 4*b^5)*cos(d*x + c)^4 + (3*a^3*b^2 - 4*a*b^4)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) +
b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 + (a^5*b - 5*a^3*b^3 + 4*
a*b^5)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) + 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 + (a^5*b - 5*a^
3*b^3 + 4*a*b^5)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(a^6 - a^4*b^2 - (a^5*b - 13*a^3*b^3 + 12*a*b^5)*c
os(d*x + c)^3 - (a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c)^2 - 2*(a^5*b - a^3*b^3)*cos(d*x + c))*sin(d*x + c))
/((a^7*b - a^5*b^3)*d*cos(d*x + c)^4 + (a^8 - a^6*b^2)*d*cos(d*x + c)^3)]

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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((1-cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.50607, size = 427, normalized size = 2.19 \begin{align*} \frac{\frac{6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} a^{4}} + \frac{3 \,{\left (a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac{3 \,{\left (a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac{6 \,{\left (3 \, a^{2} b^{2} - 4 \, b^{4}\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^{5}} - \frac{2 \,{\left (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} + 4 \, 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 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((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(6*b^3*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)*a^4) + 3*(a^2*b
- 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 3*(a^2*b - 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 +
6*(3*a^2*b^2 - 4*b^4)*(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^5) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/
2*d*x + 1/2*c)^5 + 4*a^2*tan(1/2*d*x + 1/2*c)^3 - 18*b^2*tan(1/2*d*x + 1/2*c)^3 - 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