\(\int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1345]

Optimal result

Integrand size = 29, antiderivative size = 181 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2} d}-\frac {\left (3 a^2+2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {\left (3 a^2-b^2\right ) \sec (c+d x)}{2 a \left (a^2-b^2\right ) d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d} \]

[Out]

2*b^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(3/2)/d-1/2*(3*a^2+2*b^2)*arctanh(cos(d*x
+c))/a^3/d+b*cot(d*x+c)/a^2/d+1/2*(3*a^2-b^2)*sec(d*x+c)/a/(a^2-b^2)/d-1/2*csc(d*x+c)^2*sec(d*x+c)/a/d-b*tan(d
*x+c)/(a^2-b^2)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.17, number of steps used = 17, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {2977, 2702, 327, 213, 2700, 14, 294, 2775, 12, 2739, 632, 210} \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {b \tan (c+d x)}{a^2 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {2 b^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}-\frac {3 \text {arctanh}(\cos (c+d x))}{2 a d}+\frac {3 \sec (c+d x)}{2 a d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

[In]

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

[Out]

(2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(3/2)*d) - (3*ArcTanh[Cos[c + d*x]])
/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Cot[c + d*x])/(a^2*d) + (3*Sec[c + d*x])/(2*a*d) + (b^2*Se
c[c + d*x])/(a^3*d) - (Csc[c + d*x]^2*Sec[c + d*x])/(2*a*d) + (b^3*Sec[c + d*x]*(b - a*Sin[c + d*x]))/(a^3*(a^
2 - b^2)*d) - (b*Tan[c + d*x])/(a^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 210

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

Rule 213

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

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2702

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 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2977

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b^2 \csc (c+d x) \sec ^2(c+d x)}{a^3}-\frac {b \csc ^2(c+d x) \sec ^2(c+d x)}{a^2}+\frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a}-\frac {b^3 \sec ^2(c+d x)}{a^3 (a+b \sin (c+d x))}\right ) \, dx \\ & = \frac {\int \csc ^3(c+d x) \sec ^2(c+d x) \, dx}{a}-\frac {b \int \csc ^2(c+d x) \sec ^2(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+b \sin (c+d x)} \, dx}{a^3} \\ & = \frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}+\frac {b^3 \int \frac {b^2}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {\text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{a d}-\frac {b \text {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {b^2 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d} \\ & = \frac {b^2 \sec (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}+\frac {b^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {3 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a d}-\frac {b \text {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {b^2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d} \\ & = -\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {3 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{a^2 d}+\frac {3 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a d}+\frac {\left (2 b^5\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right ) d} \\ & = -\frac {3 \text {arctanh}(\cos (c+d x))}{2 a d}-\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {3 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{a^2 d}-\frac {\left (4 b^5\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right ) d} \\ & = \frac {2 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2} d}-\frac {3 \text {arctanh}(\cos (c+d x))}{2 a d}-\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {3 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.08 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}}+\frac {4 b \cot \left (\frac {1}{2} (c+d x)\right )}{a^2}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}-\frac {4 \left (3 a^2+2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {4 \left (3 a^2+2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{a}+\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 b \tan \left (\frac {1}{2} (c+d x)\right )}{a^2}}{8 d} \]

[In]

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

[Out]

((16*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(3/2)) + (4*b*Cot[(c + d*x)/2])/a^
2 - Csc[(c + d*x)/2]^2/a - (4*(3*a^2 + 2*b^2)*Log[Cos[(c + d*x)/2]])/a^3 + (4*(3*a^2 + 2*b^2)*Log[Sin[(c + d*x
)/2]])/a^3 + Sec[(c + d*x)/2]^2/a + (8*Sin[(c + d*x)/2])/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (8*
Sin[(c + d*x)/2])/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (4*b*Tan[(c + d*x)/2])/a^2)/(8*d)

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(199\)
default	\(\frac {\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(199\)
risch	\(\frac {i \left (3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-i b^{2} a \,{\mathrm e}^{5 i \left (d x +c \right )}-2 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 i b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-4 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} b -2 b^{3}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{3} d}-\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}\)	\(437\)

[In]

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

[Out]

1/d*(1/4/a^2*(1/2*tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c))-1/8/a/tan(1/2*d*x+1/2*c)^2+1/4/a^3*(6*a^2+4*b
^2)*ln(tan(1/2*d*x+1/2*c))+1/2*b/a^2/tan(1/2*d*x+1/2*c)+2/a^3/(a-b)/(a+b)*b^5/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*
tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+1/(a-b)/(tan(1/2*d*x+1/2*c)+1)-1/(a+b)/(tan(1/2*d*x+1/2*c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (170) = 340\).

Time = 0.76 (sec) , antiderivative size = 878, normalized size of antiderivative = 4.85 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {4 \, a^{6} - 4 \, a^{4} b^{2} - 2 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (b^{5} \cos \left (d x + c\right )^{3} - b^{5} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + {\left ({\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, {\left (a^{5} b - a^{3} b^{3} - {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{3} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {4 \, a^{6} - 4 \, a^{4} b^{2} - 2 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (b^{5} \cos \left (d x + c\right )^{3} - b^{5} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left ({\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, {\left (a^{5} b - a^{3} b^{3} - {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{3} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

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

[Out]

[-1/4*(4*a^6 - 4*a^4*b^2 - 2*(3*a^6 - 4*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 - 2*(b^5*cos(d*x + c)^3 - b^5*cos(d*
x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x +
c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + (
(3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log
(1/2*cos(d*x + c) + 1/2) - ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^
4 + 2*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^5*b - a^3*b^3 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos
(d*x + c)^2)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c)^3 - (a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d
*x + c)), -1/4*(4*a^6 - 4*a^4*b^2 - 2*(3*a^6 - 4*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + 4*(b^5*cos(d*x + c)^3 - b
^5*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + ((3*a^6 - 4*a^
4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log(1/2*cos(d*x
+ c) + 1/2) - ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*co
s(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^5*b - a^3*b^3 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c)^2)*
sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c)^3 - (a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {16 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {4 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

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

[Out]

1/8*(16*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*b^5/(
(a^5 - a^3*b^2)*sqrt(a^2 - b^2)) + 16*(b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1))
+ (a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(3*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))
/a^3 - (18*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

Mupad [B] (verification not implemented)

Time = 16.10 (sec) , antiderivative size = 1570, normalized size of antiderivative = 8.67 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

(a^5*((3*b^3*sin(c + d*x))/4 - (5*b^3*sin(3*c + 3*d*x))/4) - a^7*((b*sin(c + d*x))/2 - (b*sin(3*c + 3*d*x))/2)
+ a^2*(b^6/4 + (b^6*cos(2*c + 2*d*x))/4 + (3*b^6*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 -
(3*b^6*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8) - a*((b^7*sin(c + d*x))/4 + (b^7*sin(3
*c + 3*d*x))/4) - a^6*((b^2*cos(c + d*x))/2 + b^2/4 - (7*b^2*cos(2*c + 2*d*x))/4 - (b^2*cos(3*c + 3*d*x))/2 +
(7*b^2*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (7*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d
*x)/2))*cos(3*c + 3*d*x))/8) - a^4*(b^4/4 - (b^4*cos(c + d*x))/4 + (5*b^4*cos(2*c + 2*d*x))/4 + (b^4*cos(3*c +
3*d*x))/4 - (3*b^4*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (3*b^4*log(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8) + a^8*(cos(c + d*x)/4 - (3*cos(2*c + 2*d*x))/4 - cos(3*c + 3*d*x)/4
+ (3*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
)*cos(3*c + 3*d*x))/8 + 1/4) - (b^8*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (b^8*log(sin(
c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + a^3*b^5*sin(3*c + 3*d*x) + (b^5*atan((3*a^6*sin(c/2 +
(d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 8*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*
b^2)^(1/2) + a^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 7*a^4*b^2*sin(c/2 + (d*x)/
2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 4*a*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^
(1/2) - 3*a^5*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^9*cos(c/2 + (d*x)/2)*3i - b^9
*sin(c/2 + (d*x)/2)*8i - a*b^8*cos(c/2 + (d*x)/2)*4i + a^8*b*sin(c/2 + (d*x)/2)*6i + a^3*b^6*cos(c/2 + (d*x)/2
)*5i + a^5*b^4*cos(c/2 + (d*x)/2)*3i - a^7*b^2*cos(c/2 + (d*x)/2)*7i + a^2*b^7*sin(c/2 + (d*x)/2)*12i + a^4*b^
5*sin(c/2 + (d*x)/2)*4i - a^6*b^3*sin(c/2 + (d*x)/2)*14i))*cos(3*c + 3*d*x)*(-(a + b)^3*(a - b)^3)^(1/2)*1i)/2
- (b^5*atan((3*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 8*b^6*sin(c/2 + (d*x)/2)*(b
^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)
- 7*a^4*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 4*a*b^5*cos(c/2 + (d*x)/2)*(b^6 -
a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 3*a^5*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^
9*cos(c/2 + (d*x)/2)*3i - b^9*sin(c/2 + (d*x)/2)*8i - a*b^8*cos(c/2 + (d*x)/2)*4i + a^8*b*sin(c/2 + (d*x)/2)*6
i + a^3*b^6*cos(c/2 + (d*x)/2)*5i + a^5*b^4*cos(c/2 + (d*x)/2)*3i - a^7*b^2*cos(c/2 + (d*x)/2)*7i + a^2*b^7*si
n(c/2 + (d*x)/2)*12i + a^4*b^5*sin(c/2 + (d*x)/2)*4i - a^6*b^3*sin(c/2 + (d*x)/2)*14i))*cos(c + d*x)*(-(a + b)
^3*(a - b)^3)^(1/2)*1i)/2)/(a^3*d*cos(c + d*x)*sin(c + d*x)^2*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2))