3.152 \(\int \csc ^3(a+b x) \sec (a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac {\log (\tan (a+b x))}{b}-\frac {\cot ^2(a+b x)}{2 b} \]

[Out]

-1/2*cot(b*x+a)^2/b+ln(tan(b*x+a))/b

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2620, 14} \[ \frac {\log (\tan (a+b x))}{b}-\frac {\cot ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x],x]

[Out]

-Cot[a + b*x]^2/(2*b) + Log[Tan[a + b*x]]/b

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.26 \[ -\frac {\csc ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sec[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.44, size = 65, normalized size = 2.41 \[ -\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) - {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 1}{2 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 119, normalized size = 4.41 \[ -\frac {\frac {{\left (\frac {4 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 4 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 8 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1)
/(cos(b*x + a) + 1) - 4*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 8*log(abs(-(cos(b*x + a) - 1)/(cos
(b*x + a) + 1) - 1)))/b

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maple [A]  time = 0.03, size = 26, normalized size = 0.96 \[ -\frac {1}{2 \sin \left (b x +a \right )^{2} b}+\frac {\ln \left (\tan \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 36, normalized size = 1.33 \[ -\frac {\frac {1}{\sin \left (b x + a\right )^{2}} + \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 34, normalized size = 1.26 \[ -\frac {\ln \left (\cos \left (a+b\,x\right )\right )-\frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{2}+\frac {1}{2\,{\sin \left (a+b\,x\right )}^2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)*sin(a + b*x)^3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*x)/sin(a + b*x)**3, x)

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