\(\int \csc ^3(c+d x) (a+b \sin ^2(c+d x)) \, dx\) [67]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 40 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {(a+2 b) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3091, 3855} \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {(a+2 b) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d} \]

[In]

Int[Csc[c + d*x]^3*(a + b*Sin[c + d*x]^2),x]

[Out]

-1/2*((a + 2*b)*ArcTanh[Cos[c + d*x]])/d - (a*Cot[c + d*x]*Csc[c + d*x])/(2*d)

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
 f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cot (c+d x) \csc (c+d x)}{2 d}+\frac {1}{2} (a+2 b) \int \csc (c+d x) \, dx \\ & = -\frac {(a+2 b) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(118\) vs. \(2(40)=80\).

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.95 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {b \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {b \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d} \]

[In]

Integrate[Csc[c + d*x]^3*(a + b*Sin[c + d*x]^2),x]

[Out]

-1/8*(a*Csc[(c + d*x)/2]^2)/d - (b*Log[Cos[c/2 + (d*x)/2]])/d - (a*Log[Cos[(c + d*x)/2]])/(2*d) + (b*Log[Sin[c
/2 + (d*x)/2]])/d + (a*Log[Sin[(c + d*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.30

method result size
parallelrisch \(\frac {\left (4 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{8 d}\) \(52\)
derivativedivides \(\frac {a \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\) \(59\)
default \(\frac {a \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\) \(59\)
norman \(\frac {-\frac {a}{8 d}+\frac {a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\) \(107\)
risch \(\frac {a \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\) \(110\)

[In]

int(csc(d*x+c)^3*(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/8*((4*a+8*b)*ln(tan(1/2*d*x+1/2*c))-cot(1/2*d*x+1/2*c)^2*a+tan(1/2*d*x+1/2*c)^2*a)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (36) = 72\).

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.38 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {2 \, a \cos \left (d x + c\right ) - {\left ({\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{2} - a - 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{2} - a - 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\int \left (a + b \sin ^{2}{\left (c + d x \right )}\right ) \csc ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**3*(a+b*sin(d*x+c)**2),x)

[Out]

Integral((a + b*sin(c + d*x)**2)*csc(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {{\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, a \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (36) = 72\).

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 3.02 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (a + 2 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) + \frac {{\left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(csc(d*x+c)^3*(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(2*(a + 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + (a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c)
 + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - a*(cos(d*x + c) - 1
)/(cos(d*x + c) + 1))/d

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.05 \[ \int \csc ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {a\,\cos \left (c+d\,x\right )}{2\,d\,\left ({\cos \left (c+d\,x\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )\,\left (\frac {a}{2}+b\right )}{d} \]

[In]

int((a + b*sin(c + d*x)^2)/sin(c + d*x)^3,x)

[Out]

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