3.129 \(\int \csc (a+b x) \sec ^4(a+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac {\sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b}-\frac {\tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2622, 302, 207} \[ \frac {\sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b}-\frac {\tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[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 207

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

Rule 302

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 2622

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 1.50 \[ \frac {\sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.45, size = 67, normalized size = 1.76 \[ -\frac {3 \, \cos \left (b x + a\right )^{3} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, \cos \left (b x + a\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 6 \, \cos \left (b x + a\right )^{2} - 2}{6 \, b \cos \left (b x + a\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.21, size = 101, normalized size = 2.66 \[ \frac {\frac {8 \, {\left (\frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 2\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}} + 3 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 47, normalized size = 1.24 \[ \frac {1}{3 b \cos \left (b x +a \right )^{3}}+\frac {1}{b \cos \left (b x +a \right )}+\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 50, normalized size = 1.32 \[ \frac {\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{2} + 1\right )}}{\cos \left (b x + a\right )^{3}} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.39, size = 33, normalized size = 0.87 \[ -\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )-\frac {{\cos \left (a+b\,x\right )}^2+\frac {1}{3}}{{\cos \left (a+b\,x\right )}^3}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (a + b x \right )}}{\sin {\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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