\(\int \frac {(1-\cos ^2(c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) [600]

Optimal result

Integrand size = 33, antiderivative size = 155 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \sqrt {a-b} b^2 \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d}+\frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d} \]

[Out]

1/2*b*(a^2-2*b^2)*arctanh(sin(d*x+c))/a^4/d-2*b^2*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^4/d-1/3*(a^2-3*b^2)*tan(d*x+c)/a^3/d-1/2*b*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*sec(d*x+c)^2*tan(d
*x+c)/a/d

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3135, 3134, 3080, 3855, 2738, 211} \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b^2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d}-\frac {b \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d} \]

[In]

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

[Out]

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

Rule 211

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

Rule 2738

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 3080

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 3134

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] + D
ist[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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3135

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)),
Int[(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] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (-3 b-a \cos (c+d x)+2 b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a} \\ & = -\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (-2 \left (a^2-3 b^2\right )+a b \cos (c+d x)-3 b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^2} \\ & = -\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\int \frac {\left (3 b \left (a^2-2 b^2\right )-3 a b^2 \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3} \\ & = -\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (b \left (a^2-2 b^2\right )\right ) \int \sec (c+d x) \, dx}{2 a^4}-\frac {\left (b^2 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^4} \\ & = \frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac {\left (2 b^2 \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 d} \\ & = -\frac {2 \sqrt {a-b} b^2 \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d}+\frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.03 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.65 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {24 b^2 \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )+\frac {1}{2} \sec ^3(c+d x) \left (9 b \left (a^2-2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 b \left (a^2-2 b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 a \left (-a^2-3 b^2+3 a b \cos (c+d x)+\left (a^2-3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 a^4 d} \]

[In]

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

[Out]

-1/12*(24*b^2*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + (Sec[c + d*x]^3*(9*b*(a^
2 - 2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])
+ 3*b*(a^2 - 2*b^2)*Cos[3*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]]) + 4*a*(-a^2 - 3*b^2 + 3*a*b*Cos[c + d*x] + (a^2 - 3*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/2)/(a^4
*d)

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.58

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 b^{2} \left (a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\)	\(245\)
default	\(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 b^{2} \left (a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\)	\(245\)
risch	\(\frac {i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{2}+6 b^{2}\right )}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}+\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d \,a^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d \,a^{4}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\)	\(302\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.78 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {6 \, \sqrt {-a^{2} + b^{2}} b^{2} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, a^{2} b \cos \left (d x + c\right ) - 2 \, a^{3} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a^{4} d \cos \left (d x + c\right )^{3}}, -\frac {12 \, \sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a^{2} b \cos \left (d x + c\right ) - 2 \, a^{3} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a^{4} d \cos \left (d x + c\right )^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=- \int \left (- \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^4/(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 de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.72 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {12 \, {\left (a^{2} b^{2} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(a^2*b - 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*(a^2*b - 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c)
- 1))/a^4 + 12*(a^2*b^2 - 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^4) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*
b^2*tan(1/2*d*x + 1/2*c)^5 + 8*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x
+ 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d

Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.45 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )}{3\,a\,d\,{\cos \left (c+d\,x\right )}^3}-\frac {\sin \left (c+d\,x\right )}{3\,a\,d\,\cos \left (c+d\,x\right )}-\frac {2\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^4\,d}+\frac {b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^2\,d}+\frac {2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{a^4\,d}-\frac {b\,\sin \left (c+d\,x\right )}{2\,a^2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {b^2\,\sin \left (c+d\,x\right )}{a^3\,d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

sin(c + d*x)/(3*a*d*cos(c + d*x)^3) - sin(c + d*x)/(3*a*d*cos(c + d*x)) - (2*b^3*atanh(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2)))/(a^4*d) + (b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^2*d) + (2*b^2*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^4*d) - (b*sin
(c + d*x))/(2*a^2*d*cos(c + d*x)^2) + (b^2*sin(c + d*x))/(a^3*d*cos(c + d*x))