3.44 \(\int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=118 \[ \frac{6 a^2 b^2 \sec (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^4 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}-\frac{b^4 \sec (c+d x)}{d} \]

[Out]

-((a^4*ArcTanh[Cos[c + d*x]])/d) + (4*a^3*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b^3*ArcTanh[Sin[c + d*x]])/d + (6*
a^2*b^2*Sec[c + d*x])/d - (b^4*Sec[c + d*x])/d + (b^4*Sec[c + d*x]^3)/(3*d) + (2*a*b^3*Sec[c + d*x]*Tan[c + d*
x])/d

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Rubi [A]  time = 0.114219, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{6 a^2 b^2 \sec (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^4 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{2 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}-\frac{b^4 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*(a + b*Tan[c + d*x])^4,x]

[Out]

-((a^4*ArcTanh[Cos[c + d*x]])/d) + (4*a^3*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b^3*ArcTanh[Sin[c + d*x]])/d + (6*
a^2*b^2*Sec[c + d*x])/d - (b^4*Sec[c + d*x])/d + (b^4*Sec[c + d*x]^3)/(3*d) + (2*a*b^3*Sec[c + d*x]*Tan[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]

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

Mathematica [B]  time = 5.1324, size = 352, normalized size = 2.98 \[ \frac{2 b^2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\left (36 a^2-5 b^2\right ) \cos (2 (c+d x))+36 a^2+2 b^2 \cos (c+d x)-b^2\right )+72 a^2 b^2-48 a^3 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+48 a^3 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 a^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-12 a^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 a b^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{12 a b^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+24 a b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-24 a b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b^4}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^4}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-10 b^4}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*(a + b*Tan[c + d*x])^4,x]

[Out]

(72*a^2*b^2 - 10*b^4 - 12*a^4*Log[Cos[(c + d*x)/2]] - 48*a^3*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 24*a
*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^4*Log[Sin[(c + d*x)/2]] + 48*a^3*b*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] - 24*a*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*a*b^3)/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2 + b^4/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 2*b^2*(36*a^2 - b^2 + 2*b^2*Cos[c + d*x] + (36*a
^2 - 5*b^2)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2 - (12*a*b^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^2 + b^4/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(12*d)

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Maple [A]  time = 0.069, size = 214, normalized size = 1.8 \begin{align*}{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{{b}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{2\,{b}^{4}\cos \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{b}^{3}a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{{b}^{3}a\sin \left ( dx+c \right ) }{d}}-2\,{\frac{{b}^{3}a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}}{d\cos \left ( dx+c \right ) }}+4\,{\frac{b{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{4}\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))^4,x)

[Out]

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

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Maxima [A]  time = 1.10845, size = 188, normalized size = 1.59 \begin{align*} -\frac{3 \, a 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^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 3 \, a^{4} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right ) - \frac{18 \, a^{2} b^{2}}{\cos \left (d x + c\right )} + \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{4}}{\cos \left (d x + c\right )^{3}}}{3 \, 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))^4,x, algorithm="maxima")

[Out]

-1/3*(3*a*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^3*b*
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 3*a^4*log(cot(d*x + c) + csc(d*x + c)) - 18*a^2*b^2/cos(d*x
+ c) + (3*cos(d*x + c)^2 - 1)*b^4/cos(d*x + c)^3)/d

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Fricas [A]  time = 2.56534, size = 444, normalized size = 3.76 \begin{align*} -\frac{3 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 12 \, a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b^{4} - 6 \,{\left (6 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 2.70968, size = 261, normalized size = 2.21 \begin{align*} \frac{3 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 6 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{4 \,{\left (3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, a^{2} b^{2} + b^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, 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))^4,x, algorithm="giac")

[Out]

1/3*(3*a^4*log(abs(tan(1/2*d*x + 1/2*c))) + 6*(2*a^3*b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*(2*a^3*
b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 4*(3*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*a^2*b^2*tan(1/2*d*x + 1/
2*c)^4 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*b^4*tan(1/2*d*x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - 9*a
^2*b^2 + b^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d