3.2.43 \(\int \csc ^2(a+b x) \sec ^4(a+b x) \, dx\) [143]

Optimal. Leaf size=38 \[ -\frac {\cot (a+b x)}{b}+\frac {2 \tan (a+b x)}{b}+\frac {\tan ^3(a+b x)}{3 b} \]

[Out]

-cot(b*x+a)/b+2*tan(b*x+a)/b+1/3*tan(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 = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2700, 276} \begin {gather*} \frac {\tan ^3(a+b x)}{3 b}+\frac {2 \tan (a+b x)}{b}-\frac {\cot (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sec[a + b*x]^4,x]

[Out]

-(Cot[a + b*x]/b) + (2*Tan[a + b*x])/b + Tan[a + b*x]^3/(3*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2700

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

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Mathematica [A]
time = 0.03, size = 46, normalized size = 1.21 \begin {gather*} -\frac {\cot (a+b x)}{b}+\frac {5 \tan (a+b x)}{3 b}+\frac {\sec ^2(a+b x) \tan (a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sec[a + b*x]^4,x]

[Out]

-(Cot[a + b*x]/b) + (5*Tan[a + b*x])/(3*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(3*b)

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Maple [A]
time = 0.06, size = 50, normalized size = 1.32

method	result	size
risch	\(-\frac {16 i \left (2 \,{\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}\)	\(46\)
derivativedivides	\(\frac {\frac {1}{3 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{3}}+\frac {4}{3 \sin \left (b x +a \right ) \cos \left (b x +a \right )}-\frac {8 \cot \left (b x +a \right )}{3}}{b}\)	\(50\)
default	\(\frac {\frac {1}{3 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{3}}+\frac {4}{3 \sin \left (b x +a \right ) \cos \left (b x +a \right )}-\frac {8 \cot \left (b x +a \right )}{3}}{b}\)	\(50\)
norman	\(\frac {\frac {1}{2 b}-\frac {6 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {25 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {6 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4/sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/3/sin(b*x+a)/cos(b*x+a)^3+4/3/sin(b*x+a)/cos(b*x+a)-8/3*cot(b*x+a))

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Maxima [A]
time = 0.28, size = 32, normalized size = 0.84 \begin {gather*} \frac {\tan \left (b x + a\right )^{3} - \frac {3}{\tan \left (b x + a\right )} + 6 \, \tan \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*(tan(b*x + a)^3 - 3/tan(b*x + a) + 6*tan(b*x + a))/b

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Fricas [A]
time = 0.35, size = 43, normalized size = 1.13 \begin {gather*} -\frac {8 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/3*(8*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3*sin(b*x + a))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4/sin(b*x+a)**2,x)

[Out]

Integral(sec(a + b*x)**4/sin(a + b*x)**2, x)

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Giac [A]
time = 3.97, size = 32, normalized size = 0.84 \begin {gather*} \frac {\tan \left (b x + a\right )^{3} - \frac {3}{\tan \left (b x + a\right )} + 6 \, \tan \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/3*(tan(b*x + a)^3 - 3/tan(b*x + a) + 6*tan(b*x + a))/b

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Mupad [B]
time = 0.41, size = 33, normalized size = 0.87 \begin {gather*} \frac {{\mathrm {tan}\left (a+b\,x\right )}^4+6\,{\mathrm {tan}\left (a+b\,x\right )}^2-3}{3\,b\,\mathrm {tan}\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^4*sin(a + b*x)^2),x)

[Out]

(6*tan(a + b*x)^2 + tan(a + b*x)^4 - 3)/(3*b*tan(a + b*x))

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