\(\int \csc (a+b x) \sec ^4(a+b x) \, dx\) [129]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=-\frac {\text {arctanh}(\cos (a+b x))}{b}+\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b} \]

[Out]

-arctanh(cos(b*x+a))/b+sec(b*x+a)/b+1/3*sec(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2702, 308, 213} \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=-\frac {\text {arctanh}(\cos (a+b x))}{b}+\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b} \]

[In]

Int[Csc[a + b*x]*Sec[a + b*x]^4,x]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Sec[a + b*x]/b + Sec[a + b*x]^3/(3*b)

Rule 213

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.50 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}+\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b} \]

[In]

Integrate[Csc[a + b*x]*Sec[a + b*x]^4,x]

[Out]

-(Log[Cos[(a + b*x)/2]]/b) + Log[Sin[(a + b*x)/2]]/b + Sec[a + b*x]/b + Sec[a + b*x]^3/(3*b)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {1}{3 \cos \left (b x +a \right )^{3}}+\frac {1}{\cos \left (b x +a \right )}+\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b}\)	\(40\)
default	\(\frac {\frac {1}{3 \cos \left (b x +a \right )^{3}}+\frac {1}{\cos \left (b x +a \right )}+\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b}\)	\(40\)
norman	\(\frac {\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {8}{3 b}-\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\)	\(70\)
risch	\(\frac {2 \,{\mathrm e}^{5 i \left (b x +a \right )}+\frac {20 \,{\mathrm e}^{3 i \left (b x +a \right )}}{3}+2 \,{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}\)	\(87\)
parallelrisch	\(\frac {3 \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3} \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+12 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-8}{3 b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3}}\)	\(98\)

[In]

int(sec(b*x+a)^4/sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b*(1/3/cos(b*x+a)^3+1/cos(b*x+a)+ln(csc(b*x+a)-cot(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{3} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, \cos \left (b x + a\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 6 \, \cos \left (b x + a\right )^{2} - 2}{6 \, b \cos \left (b x + a\right )^{3}} \]

[In]

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

[Out]

-1/6*(3*cos(b*x + a)^3*log(1/2*cos(b*x + a) + 1/2) - 3*cos(b*x + a)^3*log(-1/2*cos(b*x + a) + 1/2) - 6*cos(b*x
+ a)^2 - 2)/(b*cos(b*x + a)^3)

Sympy [F]

\[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=\int \frac {\sec ^{4}{\left (a + b x \right )}}{\sin {\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(sec(a + b*x)**4/sin(a + b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{2} + 1\right )}}{\cos \left (b x + a\right )^{3}} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{6 \, b} \]

[In]

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

[Out]

1/6*(2*(3*cos(b*x + a)^2 + 1)/cos(b*x + a)^3 - 3*log(cos(b*x + a) + 1) + 3*log(cos(b*x + a) - 1))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (36) = 72\).

Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.66 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {8 \, {\left (\frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 2\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}} + 3 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{6 \, b} \]

[In]

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

[Out]

1/6*(8*(3*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 2)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) + 1)^3 + 3*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \csc (a+b x) \sec ^4(a+b x) \, dx=-\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )-\frac {{\cos \left (a+b\,x\right )}^2+\frac {1}{3}}{{\cos \left (a+b\,x\right )}^3}}{b} \]

[In]

int(1/(cos(a + b*x)^4*sin(a + b*x)),x)

[Out]

-(atanh(cos(a + b*x)) - (cos(a + b*x)^2 + 1/3)/cos(a + b*x)^3)/b