3.58 \(\int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=122 \[ -\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b \csc (c+d x)}{a^2 d}+\frac {b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3518, 3110, 3768, 3770, 2621, 321, 207, 2622, 3104, 3074, 206} \[ -\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{a^3 d}+\frac {b \csc (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d
*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(a^3*d) + (b*Csc[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*
d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 321

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 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])

Rule 2622

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])

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.81, size = 179, normalized size = 1.47 \[ \frac {-16 b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )+a^2 \left (-\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 a b \tan \left (\frac {1}{2} (c+d x)\right )+4 a b \cot \left (\frac {1}{2} (c+d x)\right )+8 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + 4*a*b*Cot[(c + d*x)/2] - a^2*Csc[(
c + d*x)/2]^2 - 4*a^2*Log[Cos[(c + d*x)/2]] - 8*b^2*Log[Cos[(c + d*x)/2]] + 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^
2*Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x)/2]^2 + 4*a*b*Tan[(c + d*x)/2])/(8*a^3*d)

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fricas [B]  time = 0.53, size = 270, normalized size = 2.21 \[ \frac {2 \, a^{2} \cos \left (d x + c\right ) - 4 \, a b \sin \left (d x + c\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.07, size = 209, normalized size = 1.71 \[ \frac {\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {4 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {8 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*tan(1/2*d*x + 1/2*c)^2 + 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))
)/a^3 + 8*(a^2*b + b^3)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*
c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/
2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d

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maple [A]  time = 0.41, size = 162, normalized size = 1.33 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {1}{8 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+b*tan(d*x+c)),x)

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2+1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-1/8/d/a/tan(1/2*d*x+1/2*c)^2+1/2/d/a*ln(tan(1/2*d*
x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tan(1/2*d*x+1/2*c)-2/d*b*(a^2+b^2)^(1/2)/a^3*arctanh(
1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))

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maxima [A]  time = 0.66, size = 215, normalized size = 1.76 \[ \frac {\frac {\frac {4 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {4 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (a - \frac {4 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{2} \sin \left (d x + c\right )^{2}} + \frac {8 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((4*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 4*(a^2 + 2*b^2)*log(s
in(d*x + c)/(cos(d*x + c) + 1))/a^3 - (a - 4*b*sin(d*x + c)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)^2/(a^2*sin(
d*x + c)^2) + 8*(a^2*b + b^3)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c
)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3))/d

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mupad [B]  time = 4.45, size = 764, normalized size = 6.26 \[ \frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,\left (\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}-\frac {a^2\,\left (\frac {\cos \left (c+d\,x\right )}{2}-\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}+\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{4}\right )}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}}+\frac {a\,b\,\sin \left (c+d\,x\right )}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}}-\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2\,\left (\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}+\frac {b\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,3{}\mathrm {i}+a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,8{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2+12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {a^2+b^2}\,1{}\mathrm {i}}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,3{}\mathrm {i}+a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}\,8{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2+12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {a^2+b^2}\,1{}\mathrm {i}}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + b*tan(c + d*x))),x)

[Out]

(b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2)) - (a^2*(cos(c +
d*x)/2 - log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/4 + (log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c +
2*d*x))/4))/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) + (a*b*sin(c + d*x))/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x)
)/2) - (b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/(2*((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*
x))/2)) + (b*atan((a^4*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*8i +
a*b^3*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*4i + a^3*b*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*3i + a^2*b^2*sin(c
/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*8i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*
x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) + 5*a^3*b^2*cos(c/2 + (d*x)/2) + 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(a^2 + b^2
)^(1/2)*1i)/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) - (b*cos(2*c + 2*d*x)*atan((a^4*sin(c/2 + (d*x)/2)*(a^2 +
b^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*4i +
a^3*b*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*3i + a^2*b^2*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*8i)/(a^5*cos(c/2
+ (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) + 5*a^3*b^2*c
os(c/2 + (d*x)/2) + 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(a^2 + b^2)^(1/2)*1i)/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x)
)/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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