3.2.53 \(\int \csc ^3(a+b x) \sec ^2(a+b x) \, dx\) [153]

Optimal. Leaf size=49 \[ -\frac {3 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac {3 \sec (a+b x)}{2 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2702, 294, 327, 213} \begin {gather*} \frac {3 \sec (a+b x)}{2 b}-\frac {3 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(-3*ArcTanh[Cos[a + b*x]])/(2*b) + (3*Sec[a + b*x])/(2*b) - (Csc[a + b*x]^2*Sec[a + b*x])/(2*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 294

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(49)=98\).
time = 0.18, size = 143, normalized size = 2.92 \begin {gather*} \frac {\csc ^4(a+b x) \left (2-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-2-3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{2 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Co
s[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))
/(2*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))

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Maple [A]
time = 0.07, size = 52, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {3}{2 \cos \left (b x +a \right )}+\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(52\)
default	\(\frac {-\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {3}{2 \cos \left (b x +a \right )}+\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(52\)
norman	\(\frac {\frac {1}{8 b}+\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}-\frac {9 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}\)	\(82\)
risch	\(\frac {3 \,{\mathrm e}^{5 i \left (b x +a \right )}-2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2/sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/2/sin(b*x+a)^2/cos(b*x+a)+3/2/cos(b*x+a)+3/2*ln(csc(b*x+a)-cot(b*x+a)))

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Maxima [A]
time = 0.29, size = 61, normalized size = 1.24 \begin {gather*} \frac {\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (43) = 86\).
time = 0.38, size = 96, normalized size = 1.96 \begin {gather*} \frac {6 \, \cos \left (b x + a\right )^{2} - 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{4 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - co
s(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (a + b x \right )}}{\sin ^{3}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2/sin(b*x+a)**3,x)

[Out]

Integral(sec(a + b*x)**2/sin(a + b*x)**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (43) = 86\).
time = 3.68, size = 140, normalized size = 2.86 \begin {gather*} \frac {\frac {\frac {14 \, {\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}} + 1}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 6 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*((14*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2) - (cos(b*x + a) - 1)/(cos(b*x + a) + 1
) + 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

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Mupad [B]
time = 0.03, size = 49, normalized size = 1.00 \begin {gather*} -\frac {3\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{2\,b}-\frac {\frac {3\,{\cos \left (a+b\,x\right )}^2}{2}-1}{b\,\left (\cos \left (a+b\,x\right )-{\cos \left (a+b\,x\right )}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^2*sin(a + b*x)^3),x)

[Out]

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

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