\(\int \csc ^4(a+b x) \sec (a+b x) \, dx\) [166]

Optimal result

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

[Out]

arctanh(sin(b*x+a))/b-csc(b*x+a)/b-1/3*csc(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 = {2701, 308, 213} \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\csc ^3(a+b x)}{3 b}-\frac {\csc (a+b x)}{b} \]

[In]

Int[Csc[a + b*x]^4*Sec[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Csc[a + b*x]/b - Csc[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 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(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,\csc (a+b x)\right )}{b} \\ & = -\frac {\text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{b} \\ & = -\frac {\csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}-\frac {\text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{b} \\ & = \frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\sin ^2(a+b x)\right )}{3 b} \]

[In]

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

[Out]

-1/3*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Sin[a + b*x]^2])/b

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3}}-\frac {1}{\sin \left (b x +a \right )}+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(40\)
default	\(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3}}-\frac {1}{\sin \left (b x +a \right )}+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(40\)
parallelrisch	\(\frac {-\left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\cot ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-15 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )-15 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )+24 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-24 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{24 b}\)	\(83\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{5 i \left (b x +a \right )}-10 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{b}\)	\(90\)
norman	\(\frac {-\frac {1}{24 b}-\frac {5 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}\)	\(101\)

[In]

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

[Out]

1/b*(-1/3/sin(b*x+a)^3-1/sin(b*x+a)+ln(sec(b*x+a)+tan(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (36) = 72\).

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.47 \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=\frac {3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 8}{6 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{6 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37 \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{6 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \csc ^4(a+b x) \sec (a+b x) \, dx=\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )-\frac {{\sin \left (a+b\,x\right )}^2+\frac {1}{3}}{{\sin \left (a+b\,x\right )}^3}}{b} \]

[In]

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

[Out]

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