\(\int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [810]

Optimal result

Integrand size = 29, antiderivative size = 125 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {7 a^2 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {16 a^2 \cot (c+d x)}{3 d}-\frac {7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

[Out]

-7/2*a^2*arctanh(cos(d*x+c))/d-16/3*a^2*cot(d*x+c)/d-7/2*a^2*cot(d*x+c)*csc(d*x+c)/d+8/3*a^2*cot(d*x+c)*csc(d*
x+c)/d/(1-sin(d*x+c))+1/3*a^4*cot(d*x+c)*csc(d*x+c)/d/(a-a*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2948, 2845, 3057, 2827, 3853, 3855, 3852, 8} \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {7 a^2 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {16 a^2 \cot (c+d x)}{3 d}-\frac {7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))} \]

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(-7*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (16*a^2*Cot[c + d*x])/(3*d) - (7*a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d) +
(8*a^2*Cot[c + d*x]*Csc[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cot[c + d*x]*Csc[c + d*x])/(3*d*(a - a*Sin[
c + d*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2948

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

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 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}& = a^4 \int \frac {\csc ^3(c+d x)}{(a-a \sin (c+d x))^2} \, dx \\ & = \frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\csc ^3(c+d x) (5 a+3 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx \\ & = \frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} \int \csc ^3(c+d x) \left (21 a^2+16 a^2 \sin (c+d x)\right ) \, dx \\ & = \frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} \left (16 a^2\right ) \int \csc ^2(c+d x) \, dx+\left (7 a^2\right ) \int \csc ^3(c+d x) \, dx \\ & = -\frac {7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{2} \left (7 a^2\right ) \int \csc (c+d x) \, dx-\frac {\left (16 a^2\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d} \\ & = -\frac {7 a^2 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {16 a^2 \cot (c+d x)}{3 d}-\frac {7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.52 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-24 \cot \left (\frac {1}{2} (c+d x)\right )-3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-84 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+84 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {8}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {16 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {160 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+24 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d} \]

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(-24*Cot[(c + d*x)/2] - 3*Csc[(c + d*x)/2]^2 - 84*Log[Cos[(c + d*x)/2]] + 84*Log[Sin[(c + d*x)/2]] + 3*Se
c[(c + d*x)/2]^2 + 8/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^3 + (160*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 24*Tan[(c + d*x)/2]))/(24*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {\left (28 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-112 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {302}{3}\right ) a^{2}}{8 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(119\)
risch	\(\frac {a^{2} \left (-63 i {\mathrm e}^{5 i \left (d x +c \right )}+21 \,{\mathrm e}^{6 i \left (d x +c \right )}+126 i {\mathrm e}^{3 i \left (d x +c \right )}-98 \,{\mathrm e}^{4 i \left (d x +c \right )}-75 i {\mathrm e}^{i \left (d x +c \right )}+97 \,{\mathrm e}^{2 i \left (d x +c \right )}-32\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\)	\(148\)
derivativedivides	\(\frac {a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}-\frac {5}{6 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {5}{2 \cos \left (d x +c \right )}+\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(164\)
default	\(\frac {a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}-\frac {5}{6 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {5}{2 \cos \left (d x +c \right )}+\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(164\)
norman	\(\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2}}{8 d}-\frac {10 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {19 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {28 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {19 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {10 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {21 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {35 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {63 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {91 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {175 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(314\)

[In]

int(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/8*(28*(tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))+tan(1/2*d*x+1/2*c)^5+5*tan(1/2*d*x+1/2*c)^4+cot(1/2*d*
x+1/2*c)^2-112*tan(1/2*d*x+1/2*c)^2+5*cot(1/2*d*x+1/2*c)+190*tan(1/2*d*x+1/2*c)-302/3)*a^2/d/(tan(1/2*d*x+1/2*
c)-1)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (114) = 228\).

Time = 0.27 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.42 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {64 \, a^{2} \cos \left (d x + c\right )^{4} + 86 \, a^{2} \cos \left (d x + c\right )^{3} - 54 \, a^{2} \cos \left (d x + c\right )^{2} - 80 \, a^{2} \cos \left (d x + c\right ) - 4 \, a^{2} + 21 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) + 2 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 21 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) + 2 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (32 \, a^{2} \cos \left (d x + c\right )^{3} - 11 \, a^{2} \cos \left (d x + c\right )^{2} - 38 \, a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{3} - 3 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(64*a^2*cos(d*x + c)^4 + 86*a^2*cos(d*x + c)^3 - 54*a^2*cos(d*x + c)^2 - 80*a^2*cos(d*x + c) - 4*a^2 + 2
1*(a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) + 2*a^2 + (a^2*cos(d*x +
c)^3 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 21*(a^2*co
s(d*x + c)^4 - a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) + 2*a^2 + (a^2*cos(d*x + c)^3 + 2*
a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(32*a^2*cos(d*x
+ c)^3 - 11*a^2*cos(d*x + c)^2 - 38*a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 - d*cos(d*x + c)
^3 - 3*d*cos(d*x + c)^2 + d*cos(d*x + c) + (d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c) - 2*d)*sin(
d*x + c) + 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.28 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {8 \, {\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2} {\left (\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(8*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a^2 + a^2*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2
- 2)/(cos(d*x + c)^5 - cos(d*x + c)^3) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)) + 2*a^2*(2*(3*c
os(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.20 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 84 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (42 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {16 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{24 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*a^2*tan(1/2*d*x + 1/2*c)^2 + 84*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 24*a^2*tan(1/2*d*x + 1/2*c) - 3*(
42*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^2 - 16*(12*a^2*tan(1/2*
d*x + 1/2*c)^2 - 21*a^2*tan(1/2*d*x + 1/2*c) + 11*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.46 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {7\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {-36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {135\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {239\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a^2}{2}}{d\,\left (-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \]

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^4*sin(c + d*x)^3),x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (7*a^2*log(tan(c/2 + (d*x)/2)))/(2*d) - ((135*a^2*tan(c/2 + (d*x)/2)^3)/2 -
(239*a^2*tan(c/2 + (d*x)/2)^2)/6 - 36*a^2*tan(c/2 + (d*x)/2)^4 + a^2/2 + (5*a^2*tan(c/2 + (d*x)/2))/2)/(d*(4*
tan(c/2 + (d*x)/2)^2 - 12*tan(c/2 + (d*x)/2)^3 + 12*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^5)) + (a^2*tan
(c/2 + (d*x)/2))/d