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

Optimal. Leaf size=271 \[ \frac{b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 d \left (a^2-b^2\right )}-\frac{\left (a^2-12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{\left (3 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\tan (c+d x) \sec (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

[Out]

(b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a + b
)^(3/2)*d) - ((a^2 - 12*b^2)*ArcTanh[Sin[c + d*x]])/(2*a^5*d) - (b*(11*a^2 - 12*b^2)*Tan[c + d*x])/(2*a^4*(a^2
- b^2)*d) + ((5*a^2 - 6*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*(a^2 - b^2)*d) - (Sec[c + d*x]*Tan[c + d*x])/(
2*a*d*(a + b*Cos[c + d*x])^2) - ((3*a^2 - 4*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Cos[c
+ d*x]))

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Rubi [A]  time = 1.04766, antiderivative size = 271, 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} \[ \frac{b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 d \left (a^2-b^2\right )}-\frac{\left (a^2-12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{\left (3 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\tan (c+d x) \sec (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a + b
)^(3/2)*d) - ((a^2 - 12*b^2)*ArcTanh[Sin[c + d*x]])/(2*a^5*d) - (b*(11*a^2 - 12*b^2)*Tan[c + d*x])/(2*a^4*(a^2
- b^2)*d) + ((5*a^2 - 6*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*(a^2 - b^2)*d) - (Sec[c + d*x]*Tan[c + d*x])/(
2*a*d*(a + b*Cos[c + d*x])^2) - ((3*a^2 - 4*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*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 ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}+\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 ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 \left (5 a^4-11 a^2 b^2+6 b^4\right )-a b \left (a^2-b^2\right ) \cos (c+d x)-2 \left (3 a^2-4 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (5 a^2-6 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 b \left (11 a^4-23 a^2 b^2+12 b^4\right )-2 a \left (a^4-3 a^2 b^2+2 b^4\right ) \cos (c+d x)+2 b \left (5 a^4-11 a^2 b^2+6 b^4\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{4 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}+\frac{\left (5 a^2-6 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 \left (a^2-12 b^2\right ) \left (a^2-b^2\right )^2+2 a b \left (5 a^4-11 a^2 b^2+6 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{4 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}+\frac{\left (5 a^2-6 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (a^2-12 b^2\right ) \int \sec (c+d x) \, dx}{2 a^5}+\frac{\left (b \left (6 a^4-19 a^2 b^2+12 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2-12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}+\frac{\left (5 a^2-6 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (b \left (6 a^4-19 a^2 b^2+12 b^4\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 \left (a^2-b^2\right ) d}\\ &=\frac{b \left (6 a^4-19 a^2 b^2+12 b^4\right ) \tan ^{-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 (a^2-12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}-\frac{b \left (11 a^2-12 b^2\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right ) d}+\frac{\left (5 a^2-6 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (3 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.207, size = 414, normalized size = 1.53 \[ -\frac{b^2 \sin (c+d x)}{2 a^3 d (a+b \cos (c+d x))^2}+\frac{6 b^4 \sin (c+d x)-5 a^2 b^2 \sin (c+d x)}{2 a^4 d (a-b) (a+b) (a+b \cos (c+d x))}-\frac{b \left (-19 a^2 b^2+6 a^4+12 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^5 d \left (a^2-b^2\right ) \sqrt{b^2-a^2}}+\frac{\left (a^2-12 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac{\left (12 b^2-a^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}-\frac{3 b \sin \left (\frac{1}{2} (c+d x)\right )}{a^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{3 b \sin \left (\frac{1}{2} (c+d x)\right )}{a^4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{4 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{4 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(6*a^4 - 19*a^2*b^2 + 12*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(a^5*(a^2 - b^2)*Sqrt
[-a^2 + b^2]*d)) + ((a^2 - 12*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*a^5*d) + ((-a^2 + 12*b^2)*Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2*a^5*d) + 1/(4*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (3*b*S
in[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 1/(4*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^2) - (3*b*Sin[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (b^2*Sin[c + d*x])/(2*a^3*d*(a
+ b*Cos[c + d*x])^2) + (-5*a^2*b^2*Sin[c + d*x] + 6*b^4*Sin[c + d*x])/(2*a^4*(a - b)*(a + b)*d*(a + b*Cos[c +
d*x]))

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Maple [B]  time = 0.076, size = 747, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

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))^3,x)

[Out]

-6/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3*b^2-1/d/a^3/(a*tan(1
/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a+b)*tan(1/2*d*x+1/2*c)^3+6/d*b^4/a^4/(a*tan(1/2*d*x+1/2*c)
^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3-6/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)
^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*c)*b^2+1/d/a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a-
b)*tan(1/2*d*x+1/2*c)+6/d*b^4/a^4/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*
c)+6/d/a*b/(a^2-b^2)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-19/d/a^3*b^3/(a^
2-b^2)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+12/d*b^5/a^5/(a^2-b^2)/((a+b)*
(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+1/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/a^3
/(tan(1/2*d*x+1/2*c)-1)+3/d/a^4/(tan(1/2*d*x+1/2*c)-1)*b+1/2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)-6/d/a^5*ln(tan(1/2
*d*x+1/2*c)-1)*b^2-1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)+3/d/a^4/(tan(1/2*d*x+1/
2*c)+1)*b-1/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)+6/d/a^5*ln(tan(1/2*d*x+1/2*c)+1)*b^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((1-cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.81694, size = 2919, normalized size = 10.77 \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)^3/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((6*a^4*b^3 - 19*a^2*b^5 + 12*b^7)*cos(d*x + c)^4 + 2*(6*a^5*b^2 - 19*a^3*b^4 + 12*a*b^6)*cos(d*x + c)^3
+ (6*a^6*b - 19*a^4*b^3 + 12*a^2*b^5)*cos(d*x + c)^2)*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)) - ((a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*cos(d*x + c)^4 + 2*(a^7*b - 14*a^5*b
^3 + 25*a^3*b^5 - 12*a*b^7)*cos(d*x + c)^3 + (a^8 - 14*a^6*b^2 + 25*a^4*b^4 - 12*a^2*b^6)*cos(d*x + c)^2)*log(
sin(d*x + c) + 1) + ((a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*cos(d*x + c)^4 + 2*(a^7*b - 14*a^5*b^3 + 25*
a^3*b^5 - 12*a*b^7)*cos(d*x + c)^3 + (a^8 - 14*a^6*b^2 + 25*a^4*b^4 - 12*a^2*b^6)*cos(d*x + c)^2)*log(-sin(d*x
+ c) + 1) + 2*(a^8 - 2*a^6*b^2 + a^4*b^4 - (11*a^5*b^3 - 23*a^3*b^5 + 12*a*b^7)*cos(d*x + c)^3 - (17*a^6*b^2
- 35*a^4*b^4 + 18*a^2*b^6)*cos(d*x + c)^2 - 4*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9*
b^2 - 2*a^7*b^4 + a^5*b^6)*d*cos(d*x + c)^4 + 2*(a^10*b - 2*a^8*b^3 + a^6*b^5)*d*cos(d*x + c)^3 + (a^11 - 2*a^
9*b^2 + a^7*b^4)*d*cos(d*x + c)^2), 1/4*(2*((6*a^4*b^3 - 19*a^2*b^5 + 12*b^7)*cos(d*x + c)^4 + 2*(6*a^5*b^2 -
19*a^3*b^4 + 12*a*b^6)*cos(d*x + c)^3 + (6*a^6*b - 19*a^4*b^3 + 12*a^2*b^5)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*ar
ctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - ((a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*cos
(d*x + c)^4 + 2*(a^7*b - 14*a^5*b^3 + 25*a^3*b^5 - 12*a*b^7)*cos(d*x + c)^3 + (a^8 - 14*a^6*b^2 + 25*a^4*b^4 -
12*a^2*b^6)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((a^6*b^2 - 14*a^4*b^4 + 25*a^2*b^6 - 12*b^8)*cos(d*x + c
)^4 + 2*(a^7*b - 14*a^5*b^3 + 25*a^3*b^5 - 12*a*b^7)*cos(d*x + c)^3 + (a^8 - 14*a^6*b^2 + 25*a^4*b^4 - 12*a^2*
b^6)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(a^8 - 2*a^6*b^2 + a^4*b^4 - (11*a^5*b^3 - 23*a^3*b^5 + 12*a*b
^7)*cos(d*x + c)^3 - (17*a^6*b^2 - 35*a^4*b^4 + 18*a^2*b^6)*cos(d*x + c)^2 - 4*(a^7*b - 2*a^5*b^3 + a^3*b^5)*c
os(d*x + c))*sin(d*x + c))/((a^9*b^2 - 2*a^7*b^4 + a^5*b^6)*d*cos(d*x + c)^4 + 2*(a^10*b - 2*a^8*b^3 + a^6*b^5
)*d*cos(d*x + c)^3 + (a^11 - 2*a^9*b^2 + a^7*b^4)*d*cos(d*x + c)^2)]

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

[Out]

Timed out

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Giac [B]  time = 1.45675, size = 860, normalized size = 3.17 \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)^3/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(6*a^4*b - 19*a^2*b^3 + 12*b^5)*(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)) - 2*(a^5*tan(1/2*d
*x + 1/2*c)^7 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^7 - 18*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 + 7*a^2*b^3*tan(1/2*d*x + 1
/2*c)^7 + 18*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 12*b^5*tan(1/2*d*x + 1/2*c)^7 + 3*a^5*tan(1/2*d*x + 1/2*c)^5 + 4*a
^4*b*tan(1/2*d*x + 1/2*c)^5 + 14*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 37*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 18*a*b^4
*tan(1/2*d*x + 1/2*c)^5 + 36*b^5*tan(1/2*d*x + 1/2*c)^5 + 3*a^5*tan(1/2*d*x + 1/2*c)^3 - 4*a^4*b*tan(1/2*d*x +
1/2*c)^3 + 14*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 37*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 18*a*b^4*tan(1/2*d*x + 1/2
*c)^3 - 36*b^5*tan(1/2*d*x + 1/2*c)^3 + a^5*tan(1/2*d*x + 1/2*c) - 4*a^4*b*tan(1/2*d*x + 1/2*c) - 18*a^3*b^2*t
an(1/2*d*x + 1/2*c) - 7*a^2*b^3*tan(1/2*d*x + 1/2*c) + 18*a*b^4*tan(1/2*d*x + 1/2*c) + 12*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)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^2 - a -
b)^2) + (a^2 - 12*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - (a^2 - 12*b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^5)/d