3.16 \(\int \frac{\sec ^4(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=109 \[ \frac{2 a^3 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2} \]

[Out]

(2*a^3*b*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (Sec[x]^3*(b - a*Sin[x]))/(3*(a^2 - b^
2)) - (Sec[x]*(3*a^2*b - a*(2*a^2 + b^2)*Sin[x]))/(3*(a^2 - b^2)^2)

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Rubi [A]  time = 0.238833, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2866, 12, 2660, 618, 206} \[ \frac{2 a^3 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + b*Csc[x]),x]

[Out]

(2*a^3*b*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (Sec[x]^3*(b - a*Sin[x]))/(3*(a^2 - b^
2)) - (Sec[x]*(3*a^2*b - a*(2*a^2 + b^2)*Sin[x]))/(3*(a^2 - b^2)^2)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^4(x)}{a+b \csc (x)} \, dx &=\int \frac{\sec ^3(x) \tan (x)}{b+a \sin (x)} \, dx\\ &=-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}+\frac{\int \frac{\sec ^2(x) \left (-a b+2 a^2 \sin (x)\right )}{b+a \sin (x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}+\frac{\int -\frac{3 a^3 b}{b+a \sin (x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}-\frac{\left (a^3 b\right ) \int \frac{1}{b+a \sin (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}-\frac{\left (2 a^3 b\right ) \operatorname{Subst}\left (\int \frac{1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}+\frac{\left (4 a^3 b\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2-b^2\right )-x^2} \, dx,x,2 a+2 b \tan \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac{2 a^3 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.757917, size = 121, normalized size = 1.11 \[ \frac{\sec ^3(x) \left (-6 a^2 b \cos (2 x)-10 a^2 b+6 a^3 \sin (x)+2 a^3 \sin (3 x)-3 a b^2 \sin (x)+a b^2 \sin (3 x)+4 b^3\right )-\frac{24 a^3 b \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{12 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4/(a + b*Csc[x]),x]

[Out]

((-24*a^3*b*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + Sec[x]^3*(-10*a^2*b + 4*b^3 - 6*a^2*
b*Cos[2*x] + 6*a^3*Sin[x] - 3*a*b^2*Sin[x] + 2*a^3*Sin[3*x] + a*b^2*Sin[3*x]))/(12*(a - b)^2*(a + b)^2)

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Maple [B]  time = 0.061, size = 200, normalized size = 1.8 \begin{align*} -{\frac{4}{12\,a-12\,b} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{ \left ( 4\,a-4\,b \right ) \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}-{\frac{a}{ \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{2\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{4}{12\,a+12\,b} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-2\,{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( \tan \left ( x/2 \right ) -1 \right ) ^{2}}}-{\frac{b}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{ \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{a}^{3}b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4/(a+b*csc(x)),x)

[Out]

-4/3/(tan(1/2*x)+1)^3/(4*a-4*b)+2/(4*a-4*b)/(tan(1/2*x)+1)^2-1/(a-b)^2/(tan(1/2*x)+1)*a+1/2/(a-b)^2/(tan(1/2*x
)+1)*b-4/3/(tan(1/2*x)-1)^3/(4*a+4*b)-2/(4*a+4*b)/(tan(1/2*x)-1)^2-1/2/(a+b)^2/(tan(1/2*x)-1)*b-1/(a+b)^2/(tan
(1/2*x)-1)*a-2*a^3*b/(a-b)^2/(a+b)^2/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(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(x)^4/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.56328, size = 882, normalized size = 8.09 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - b^{2}} a^{3} b \cos \left (x\right )^{3} \log \left (\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \,{\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{4} b + 4 \, a^{2} b^{3} - 2 \, b^{5} - 6 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{3}}, \frac{3 \, \sqrt{-a^{2} + b^{2}} a^{3} b \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right )^{3} - a^{4} b + 2 \, a^{2} b^{3} - b^{5} - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )^{2} +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(a^2 - b^2)*a^3*b*cos(x)^3*log(((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*si
n(x) + a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*a^4*b + 4*a^2*b^3 - 2*b^5 - 6
*(a^4*b - a^2*b^3)*cos(x)^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4 + (2*a^5 - a^3*b^2 - a*b^4)*cos(x)^2)*sin(x))/((a^6 -
3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^3), 1/3*(3*sqrt(-a^2 + b^2)*a^3*b*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)
/((a^2 - b^2)*cos(x)))*cos(x)^3 - a^4*b + 2*a^2*b^3 - b^5 - 3*(a^4*b - a^2*b^3)*cos(x)^2 + (a^5 - 2*a^3*b^2 +
a*b^4 + (2*a^5 - a^3*b^2 - a*b^4)*cos(x)^2)*sin(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (x \right )}}{a + b \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**4/(a+b*csc(x)),x)

[Out]

Integral(sec(x)**4/(a + b*csc(x)), x)

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Giac [A]  time = 1.39852, size = 252, normalized size = 2.31 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{4} + 3 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{4} - 2 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 4 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} + 3 \, a^{3} \tan \left (\frac{1}{2} \, x\right ) - 4 \, a^{2} b + b^{3}\right )}}{3 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*csc(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*a^3*b/((a^4 - 2*a^2*b^2 + b
^4)*sqrt(-a^2 + b^2)) - 2/3*(3*a^3*tan(1/2*x)^5 - 6*a^2*b*tan(1/2*x)^4 + 3*b^3*tan(1/2*x)^4 - 2*a^3*tan(1/2*x)
^3 - 4*a*b^2*tan(1/2*x)^3 + 6*a^2*b*tan(1/2*x)^2 + 3*a^3*tan(1/2*x) - 4*a^2*b + b^3)/((a^4 - 2*a^2*b^2 + b^4)*
(tan(1/2*x)^2 - 1)^3)