Integrand size = 29, antiderivative size = 100 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (3 a^2+2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {2 a b \cot (c+d x)}{d}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d} \] Output:
-1/2*(3*a^2+2*b^2)*arctanh(cos(d*x+c))/d-2*a*b*cot(d*x+c)/d+1/2*(3*a^2+2*b ^2)*sec(d*x+c)/d-1/2*a^2*csc(d*x+c)^2*sec(d*x+c)/d+2*a*b*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(100)=200\).
Time = 1.73 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.38 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\csc ^4(c+d x) \left (2 a^2+4 b^2-2 \left (3 a^2+2 b^2\right ) \cos (2 (c+d x))+3 a^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\left (3 a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-3 a^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \sin (c+d x)-8 a b \sin (3 (c+d x))\right )}{2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Csc[c + d*x]^3*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]
Output:
(Csc[c + d*x]^4*(2*a^2 + 4*b^2 - 2*(3*a^2 + 2*b^2)*Cos[2*(c + d*x)] + 3*a^ 2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 2*b^2*Cos[3*(c + d*x)]*Log[Cos[ (c + d*x)/2]] - (3*a^2 + 2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2]] - Log[ Sin[(c + d*x)/2]]) - 3*a^2*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 2*b^2* Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 8*a*b*Sin[c + d*x] - 8*a*b*Sin[3* (c + d*x)]))/(2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))
Time = 0.57 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.29, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3390, 3042, 3100, 244, 2009, 3680, 354, 87, 61, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sin (c+d x)^3 \cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3390 |
| \(\displaystyle \int \csc ^3(c+d x) \sec ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )dx+2 a b \int \csc ^2(c+d x) \sec ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^3}dx+2 a b \int \csc (c+d x)^2 \sec (c+d x)^2dx\) |
| \(\Big \downarrow \) 3100 |
| \(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^3}dx+\frac {2 a b \int \cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^3}dx+\frac {2 a b \int \left (\cot ^2(c+d x)+1\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {a^2+b^2 \sin (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^3}dx+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 3680 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \int \frac {\csc ^3(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )}{\left (1-\sin ^2(c+d x)\right )^{3/2}}d\sin (c+d x)}{d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \int \frac {\csc ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )}{\left (1-\sin ^2(c+d x)\right )^{3/2}}d\sin ^2(c+d x)}{2 d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (\frac {1}{2} \left (3 a^2+2 b^2\right ) \int \frac {\csc (c+d x)}{\left (1-\sin ^2(c+d x)\right )^{3/2}}d\sin ^2(c+d x)-\frac {a^2 \csc (c+d x)}{\sqrt {1-\sin ^2(c+d x)}}\right )}{2 d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (\frac {1}{2} \left (3 a^2+2 b^2\right ) \left (\int \frac {\csc (c+d x)}{\sqrt {1-\sin ^2(c+d x)}}d\sin ^2(c+d x)+\frac {2}{\sqrt {1-\sin ^2(c+d x)}}\right )-\frac {a^2 \csc (c+d x)}{\sqrt {1-\sin ^2(c+d x)}}\right )}{2 d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (\frac {1}{2} \left (3 a^2+2 b^2\right ) \left (\frac {2}{\sqrt {1-\sin ^2(c+d x)}}-2 \int \frac {1}{1-\sin ^4(c+d x)}d\sqrt {1-\sin ^2(c+d x)}\right )-\frac {a^2 \csc (c+d x)}{\sqrt {1-\sin ^2(c+d x)}}\right )}{2 d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (\frac {1}{2} \left (3 a^2+2 b^2\right ) \left (\frac {2}{\sqrt {1-\sin ^2(c+d x)}}-2 \text {arctanh}\left (\sqrt {1-\sin ^2(c+d x)}\right )\right )-\frac {a^2 \csc (c+d x)}{\sqrt {1-\sin ^2(c+d x)}}\right )}{2 d}+\frac {2 a b (\tan (c+d x)-\cot (c+d x))}{d}\) |
Input:
Int[Csc[c + d*x]^3*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]
Output:
(Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*(-((a^2*Csc[c + d*x])/Sqrt[1 - Sin[c + d*x]^2]) + ((3*a^2 + 2*b^2)*(-2*ArcTanh[Sqrt[1 - Sin[c + d*x]^2]] + 2/Sqrt [1 - Sin[c + d*x]^2]))/2))/(2*d) + (2*a*b*(-Cot[c + d*x] + Tan[c + d*x]))/ d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFac tors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])) S ubst[Int[(d*ff*x)^n*(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, S in[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[m/2]
Time = 1.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+b^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(116\) |
| default | \(\frac {a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+b^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(116\) |
| parallelrisch | \(\frac {\left (\left (12 a^{2}+8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 a^{2}-8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a b +\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) a b -48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a b -18 a^{2}-16 b^{2}}{8 d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-8 d}\) | \(149\) |
| risch | \(\frac {3 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{i \left (d x +c \right )}-8 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+8 i a b}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}\) | \(215\) |
| norman | \(\frac {\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {a^{2}}{8 d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8 d}-\frac {\left (7 a^{2}+8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 d}-\frac {\left (19 a^{2}+16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}-\frac {\left (33 a^{2}+32 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {10 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\) | \(265\) |
Input:
int(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2/sin(d*x+c)^2/cos(d*x+c)+3/2/cos(d*x+c)+3/2*ln(csc(d*x+c)-co t(d*x+c)))+2*a*b*(1/sin(d*x+c)/cos(d*x+c)-2*cot(d*x+c))+b^2*(1/cos(d*x+c)+ ln(csc(d*x+c)-cot(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.86 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, a^{2} - 4 \, b^{2} - {\left ({\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 8 \, {\left (2 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/4*(2*(3*a^2 + 2*b^2)*cos(d*x + c)^2 - 4*a^2 - 4*b^2 - ((3*a^2 + 2*b^2)*c os(d*x + c)^3 - (3*a^2 + 2*b^2)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + ((3*a^2 + 2*b^2)*cos(d*x + c)^3 - (3*a^2 + 2*b^2)*cos(d*x + c))*log(-1/2 *cos(d*x + c) + 1/2) + 8*(2*a*b*cos(d*x + c)^2 - a*b)*sin(d*x + c))/(d*cos (d*x + c)^3 - d*cos(d*x + c))
Timed out. \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**3*sec(d*x+c)**2*(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.23 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, a b {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{4 \, d} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/4*(a^2*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*log (cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) + 2*b^2*(2/cos(d*x + c) - lo g(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 8*a*b*(1/tan(d*x + c) - tan (d*x + c)))/d
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.57 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {16 \, {\left (2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} + b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) + 4*(3*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - 16*(2*a*b*tan(1/2*d*x + 1/2*c) + a ^2 + b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (18*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2* d*x + 1/2*c)^2)/d
Time = 19.40 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.48 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{2}+b^2\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {17\,a^2}{2}+8\,b^2\right )-\frac {a^2}{2}+20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \] Input:
int((a + b*sin(c + d*x))^2/(cos(c + d*x)^2*sin(c + d*x)^3),x)
Output:
(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (log(tan(c/2 + (d*x)/2))*((3*a^2)/2 + b ^2))/d + (tan(c/2 + (d*x)/2)^2*((17*a^2)/2 + 8*b^2) - a^2/2 + 20*a*b*tan(c /2 + (d*x)/2)^3 - 4*a*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 - 4 *tan(c/2 + (d*x)/2)^4)) + (a*b*tan(c/2 + (d*x)/2))/d
Time = 0.20 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.71 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}+8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{2}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}+32 \sin \left (d x +c \right )^{3} a b +12 \sin \left (d x +c \right )^{2} a^{2}+8 \sin \left (d x +c \right )^{2} b^{2}-16 \sin \left (d x +c \right ) a b -4 a^{2}}{8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d} \] Input:
int(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x)
Output:
(12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2 + 8*cos(c + d* x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**2 - 9*cos(c + d*x)*sin(c + d*x )**2*a**2 - 8*cos(c + d*x)*sin(c + d*x)**2*b**2 + 32*sin(c + d*x)**3*a*b + 12*sin(c + d*x)**2*a**2 + 8*sin(c + d*x)**2*b**2 - 16*sin(c + d*x)*a*b - 4*a**2)/(8*cos(c + d*x)*sin(c + d*x)**2*d)