3.35 \(\int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=86 \[ \frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

-((a^3*ArcTanh[Cos[c + d*x]])/d) + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (b^3*ArcTanh[Sin[c + d*x]])/(2*d) + (3*
a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.0856801, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3517, 3770, 2606, 8, 2611} \[ \frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

-((a^3*ArcTanh[Cos[c + d*x]])/d) + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (b^3*ArcTanh[Sin[c + d*x]])/(2*d) + (3*
a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]*Tan[c + d*x])/(2*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]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rubi steps

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

Mathematica [B]  time = 2.25956, size = 241, normalized size = 2.8 \[ \frac{-12 a^2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 a^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 a b^2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)+12 a b^2+\frac{b^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+2 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

(12*a*b^2 - 4*a^3*Log[Cos[(c + d*x)/2]] - 12*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*b^3*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]] + 4*a^3*Log[Sin[(c + d*x)/2]] + 12*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]] - 2*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 24*a*b^2*S
ec[c + d*x]*Sin[(c + d*x)/2]^2 - b^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(4*d)

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Maple [A]  time = 0.061, size = 125, normalized size = 1.5 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{a{b}^{2}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{b{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}\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))^3,x)

[Out]

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

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Maxima [A]  time = 1.08484, size = 150, normalized size = 1.74 \begin{align*} -\frac{b^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{3} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right ) - \frac{12 \, a b^{2}}{\cos \left (d x + c\right )}}{4 \, 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))^3,x, algorithm="maxima")

[Out]

-1/4*(b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*a^2*b*(log
(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*a^3*log(cot(d*x + c) + csc(d*x + c)) - 12*a*b^2/cos(d*x + c))/
d

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Fricas [A]  time = 2.49657, size = 383, normalized size = 4.45 \begin{align*} -\frac{2 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 12 \, a b^{2} \cos \left (d x + c\right ) -{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b^{3} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^3*cos(d*x + c)^2*log(1/2*cos(d*x + c) + 1/2) - 2*a^3*cos(d*x + c)^2*log(-1/2*cos(d*x + c) + 1/2) - 1
2*a*b^2*cos(d*x + c) - (6*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (6*a^2*b - b^3)*cos(d*x + c)^2*l
og(-sin(d*x + c) + 1) - 2*b^3*sin(d*x + c))/(d*cos(d*x + c)^2)

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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 )^{3} \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))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*csc(c + d*x), x)

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Giac [A]  time = 2.06041, size = 194, normalized size = 2.26 \begin{align*} \frac{2 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) +{\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, 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))^3,x, algorithm="giac")

[Out]

1/2*(2*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + (6*a^2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (6*a^2*b - b^
3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d*x + 1/2*c)^2 + b^3*t
an(1/2*d*x + 1/2*c) + 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d