3.457 \(\int \frac{\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ -\frac{2 b^3 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{b \tan (c+d x)}{a^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

(-2*b^3*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 + 2*b^2)*A
rcTanh[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)

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Rubi [A]  time = 0.324242, antiderivative size = 119, normalized size of antiderivative = 1., 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 = {2802, 3055, 3001, 3770, 2659, 205} \[ -\frac{2 b^3 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{b \tan (c+d x)}{a^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^3*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 + 2*b^2)*A
rcTanh[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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*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*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((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{\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{b^3 \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^3}+\frac{\left (a^2+2 b^2\right ) \int \sec (c+d x) \, dx}{2 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^3\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 b^3 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b} 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*}

Mathematica [A]  time = 1.00293, size = 238, normalized size = 2. \[ \frac{\frac{8 b^3 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+\frac{a^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-2 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 a b \tan (c+d x)-4 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^3*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^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/(Co
s[(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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Maple [B]  time = 0.112, size = 262, normalized size = 2.2 \begin{align*}{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{{b}^{3}}{d{a}^{3}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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(ta
n(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/a/d/(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^2-2/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))

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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(sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.53718, size = 1061, normalized size = 8.92 \begin{align*} \left [-\frac{2 \, \sqrt{-a^{2} + b^{2}} b^{3} \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^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{4} - a^{2} b^{2} - 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac{4 \, \sqrt{a^{2} - b^{2}} b^{3} \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^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{4} - a^{2} b^{2} - 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

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Giac [A]  time = 1.44655, size = 285, normalized size = 2.39 \begin{align*} \frac{\frac{4 \,{\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 )} b^{3}}{\sqrt{a^{2} - b^{2}} 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{{\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{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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)))*b^3/(sqrt(a^2 - b^2)*a^3) + (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 + 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