3.26 \(\int \csc (c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=43 \[ -\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \sec (c+d x)}{d} \]

[Out]

-((a^2*ArcTanh[Cos[c + d*x]])/d) + (2*a*b*ArcTanh[Sin[c + d*x]])/d + (b^2*Sec[c + d*x])/d

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Rubi [A]  time = 0.0563928, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3517, 3770, 2606, 8} \[ -\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

-((a^2*ArcTanh[Cos[c + d*x]])/d) + (2*a*b*ArcTanh[Sin[c + d*x]])/d + (b^2*Sec[c + d*x])/d

Rule 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [B]  time = 0.233457, size = 97, normalized size = 2.26 \[ \frac{a \left (a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

(a*(-(a*Log[Cos[(c + d*x)/2]]) - 2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + a*Log[Sin[(c + d*x)/2]] + 2*b*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + b^2*Sec[c + d*x])/d

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Maple [A]  time = 0.044, size = 61, normalized size = 1.4 \begin{align*}{\frac{{b}^{2}}{d\cos \left ( dx+c \right ) }}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*(a+b*tan(d*x+c))^2,x)

[Out]

1/d*b^2/cos(d*x+c)+2/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.06914, size = 81, normalized size = 1.88 \begin{align*} \frac{a b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - a^{2} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right ) + \frac{b^{2}}{\cos \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

(a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - a^2*log(cot(d*x + c) + csc(d*x + c)) + b^2/cos(d*x + c)
)/d

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Fricas [B]  time = 2.4343, size = 288, normalized size = 6.7 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - a^{2} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b^{2}}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(a^2*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2) - a^2*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2) - 2*a*b*cos
(d*x + c)*log(sin(d*x + c) + 1) + 2*a*b*cos(d*x + c)*log(-sin(d*x + c) + 1) - 2*b^2)/(d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \csc{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*csc(c + d*x), x)

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Giac [A]  time = 1.5719, size = 100, normalized size = 2.33 \begin{align*} \frac{2 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \, b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

(2*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + a^2*log(abs(tan(1/2*d*x
+ 1/2*c))) - 2*b^2/(tan(1/2*d*x + 1/2*c)^2 - 1))/d