3.168 \(\int \csc ^4(a+b x) \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac {5 \csc ^3(a+b x)}{6 b}-\frac {5 \csc (a+b x)}{2 b}+\frac {5 \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b} \]

[Out]

5/2*arctanh(sin(b*x+a))/b-5/2*csc(b*x+a)/b-5/6*csc(b*x+a)^3/b+1/2*csc(b*x+a)^3*sec(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2621, 288, 302, 207} \[ -\frac {5 \csc ^3(a+b x)}{6 b}-\frac {5 \csc (a+b x)}{2 b}+\frac {5 \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4*Sec[a + b*x]^3,x]

[Out]

(5*ArcTanh[Sin[a + b*x]])/(2*b) - (5*Csc[a + b*x])/(2*b) - (5*Csc[a + b*x]^3)/(6*b) + (Csc[a + b*x]^3*Sec[a +
b*x]^2)/(2*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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 2621

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*} \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b}\\ &=\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{2 b}\\ &=-\frac {5 \csc (a+b x)}{2 b}-\frac {5 \csc ^3(a+b x)}{6 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b}\\ &=\frac {5 \tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {5 \csc (a+b x)}{2 b}-\frac {5 \csc ^3(a+b x)}{6 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 31, normalized size = 0.47 \[ -\frac {\csc ^3(a+b x) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\sin ^2(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^4*Sec[a + b*x]^3,x]

[Out]

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

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fricas [B]  time = 0.44, size = 130, normalized size = 1.97 \[ -\frac {30 \, \cos \left (b x + a\right )^{4} - 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{12 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/12*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b*x + a) + 1)*sin(b*x + a) + 15*(cos(b
*x + a)^4 - cos(b*x + a)^2)*log(-sin(b*x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 -
b*cos(b*x + a)^2)*sin(b*x + a))

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giac [A]  time = 0.54, size = 72, normalized size = 1.09 \[ -\frac {\frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac {4 \, {\left (6 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/12*(6*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(6*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 15*log(abs(sin(b*x + a)
+ 1)) + 15*log(abs(sin(b*x + a) - 1)))/b

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maple [A]  time = 0.04, size = 76, normalized size = 1.15 \[ -\frac {1}{3 b \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {5}{6 b \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {5}{2 b \sin \left (b x +a \right )}+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3/sin(b*x+a)^4,x)

[Out]

-1/3/b/sin(b*x+a)^3/cos(b*x+a)^2+5/6/b/sin(b*x+a)/cos(b*x+a)^2-5/2/b/sin(b*x+a)+5/2/b*ln(sec(b*x+a)+tan(b*x+a)
)

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maxima [A]  time = 0.31, size = 73, normalized size = 1.11 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{4} - 10 \, \sin \left (b x + a\right )^{2} - 2\right )}}{\sin \left (b x + a\right )^{5} - \sin \left (b x + a\right )^{3}} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/12*(2*(15*sin(b*x + a)^4 - 10*sin(b*x + a)^2 - 2)/(sin(b*x + a)^5 - sin(b*x + a)^3) - 15*log(sin(b*x + a) +
1) + 15*log(sin(b*x + a) - 1))/b

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mupad [B]  time = 0.38, size = 61, normalized size = 0.92 \[ \frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2\,b}-\frac {-\frac {5\,{\sin \left (a+b\,x\right )}^4}{2}+\frac {5\,{\sin \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\sin \left (a+b\,x\right )}^3-{\sin \left (a+b\,x\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^3*sin(a + b*x)^4),x)

[Out]

(5*atanh(sin(a + b*x)))/(2*b) - ((5*sin(a + b*x)^2)/3 - (5*sin(a + b*x)^4)/2 + 1/3)/(b*(sin(a + b*x)^3 - sin(a
+ b*x)^5))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (a + b x \right )}}{\sin ^{4}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3/sin(b*x+a)**4,x)

[Out]

Integral(sec(a + b*x)**3/sin(a + b*x)**4, x)

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