3.15 \(\int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=85 \[ -\frac {a^2 b \log (a \sin (x)+b)}{\left (a^2-b^2\right )^2}-\frac {\sec ^2(x) (b-a \sin (x))}{2 \left (a^2-b^2\right )}-\frac {a \log (1-\sin (x))}{4 (a+b)^2}+\frac {a \log (\sin (x)+1)}{4 (a-b)^2} \]

[Out]

-1/4*a*ln(1-sin(x))/(a+b)^2+1/4*a*ln(1+sin(x))/(a-b)^2-a^2*b*ln(b+a*sin(x))/(a^2-b^2)^2-1/2*sec(x)^2*(b-a*sin(
x))/(a^2-b^2)

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Rubi [A]  time = 0.19, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3872, 2837, 12, 823, 801} \[ -\frac {a^2 b \log (a \sin (x)+b)}{\left (a^2-b^2\right )^2}-\frac {\sec ^2(x) (b-a \sin (x))}{2 \left (a^2-b^2\right )}-\frac {a \log (1-\sin (x))}{4 (a+b)^2}+\frac {a \log (\sin (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Csc[x]),x]

[Out]

-(a*Log[1 - Sin[x]])/(4*(a + b)^2) + (a*Log[1 + Sin[x]])/(4*(a - b)^2) - (a^2*b*Log[b + a*Sin[x]])/(a^2 - b^2)
^2 - (Sec[x]^2*(b - a*Sin[x]))/(2*(a^2 - b^2))

Rule 12

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2837

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.41, size = 139, normalized size = 1.64 \[ \frac {\csc (x) (a \sin (x)+b) \left (-\frac {4 a^2 b \log (a \sin (x)+b)}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}+\frac {1}{(b-a) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}-\frac {2 a \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}+\frac {2 a \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )}{(a-b)^2}\right )}{4 (a+b \csc (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + b*Csc[x]),x]

[Out]

(Csc[x]*((-2*a*Log[Cos[x/2] - Sin[x/2]])/(a + b)^2 + (2*a*Log[Cos[x/2] + Sin[x/2]])/(a - b)^2 - (4*a^2*b*Log[b
+ a*Sin[x]])/(a^2 - b^2)^2 + 1/((a + b)*(Cos[x/2] - Sin[x/2])^2) + 1/((-a + b)*(Cos[x/2] + Sin[x/2])^2))*(b +
a*Sin[x]))/(4*(a + b*Csc[x]))

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fricas [A]  time = 0.73, size = 119, normalized size = 1.40 \[ -\frac {4 \, a^{2} b \cos \relax (x)^{2} \log \left (a \sin \relax (x) + b\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \relax (x)}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a^2*b*cos(x)^2*log(a*sin(x) + b) - (a^3 + 2*a^2*b + a*b^2)*cos(x)^2*log(sin(x) + 1) + (a^3 - 2*a^2*b +
a*b^2)*cos(x)^2*log(-sin(x) + 1) + 2*a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*sin(x))/((a^4 - 2*a^2*b^2 + b^4)*cos(x)^
2)

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giac [A]  time = 1.49, size = 129, normalized size = 1.52 \[ -\frac {a^{3} b \log \left ({\left | a \sin \relax (x) + b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {a \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {a \log \left (-\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a^{2} b - b^{3} - {\left (a^{3} - a b^{2}\right )} \sin \relax (x)}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\sin \relax (x) + 1\right )} {\left (\sin \relax (x) - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*b*log(abs(a*sin(x) + b))/(a^5 - 2*a^3*b^2 + a*b^4) + 1/4*a*log(sin(x) + 1)/(a^2 - 2*a*b + b^2) - 1/4*a*lo
g(-sin(x) + 1)/(a^2 + 2*a*b + b^2) + 1/2*(a^2*b - b^3 - (a^3 - a*b^2)*sin(x))/((a + b)^2*(a - b)^2*(sin(x) + 1
)*(sin(x) - 1))

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maple [A]  time = 0.51, size = 89, normalized size = 1.05 \[ -\frac {a^{2} b \ln \left (b +a \sin \relax (x )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (-1+\sin \relax (x )\right )}-\frac {a \ln \left (-1+\sin \relax (x )\right )}{4 \left (a +b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \relax (x )\right )}+\frac {a \ln \left (1+\sin \relax (x )\right )}{4 \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3/(a+b*csc(x)),x)

[Out]

-a^2*b/(a+b)^2/(a-b)^2*ln(b+a*sin(x))-1/(4*a+4*b)/(-1+sin(x))-1/4*a/(a+b)^2*ln(-1+sin(x))-1/(4*a-4*b)/(1+sin(x
))+1/4*a*ln(1+sin(x))/(a-b)^2

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maxima [A]  time = 0.32, size = 108, normalized size = 1.27 \[ -\frac {a^{2} b \log \left (a \sin \relax (x) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {a \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {a \log \left (\sin \relax (x) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a \sin \relax (x) - b}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \sin \relax (x)^{2} - a^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*csc(x)),x, algorithm="maxima")

[Out]

-a^2*b*log(a*sin(x) + b)/(a^4 - 2*a^2*b^2 + b^4) + 1/4*a*log(sin(x) + 1)/(a^2 - 2*a*b + b^2) - 1/4*a*log(sin(x
) - 1)/(a^2 + 2*a*b + b^2) - 1/2*(a*sin(x) - b)/((a^2 - b^2)*sin(x)^2 - a^2 + b^2)

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mupad [B]  time = 0.38, size = 96, normalized size = 1.13 \[ \ln \left (b+a\,\sin \relax (x)\right )\,\left (\frac {a}{4\,{\left (a+b\right )}^2}-\frac {a}{4\,{\left (a-b\right )}^2}\right )-\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \relax (x)}{2\,\left (a^2-b^2\right )}}{{\cos \relax (x)}^2}+\frac {a\,\ln \left (\sin \relax (x)+1\right )}{4\,{\left (a-b\right )}^2}-\frac {a\,\ln \left (\sin \relax (x)-1\right )}{4\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + b/sin(x))),x)

[Out]

log(b + a*sin(x))*(a/(4*(a + b)^2) - a/(4*(a - b)^2)) - (b/(2*(a^2 - b^2)) - (a*sin(x))/(2*(a^2 - b^2)))/cos(x
)^2 + (a*log(sin(x) + 1))/(4*(a - b)^2) - (a*log(sin(x) - 1))/(4*(a + b)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+b*csc(x)),x)

[Out]

Integral(sec(x)**3/(a + b*csc(x)), x)

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