\(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1499]

Optimal result

Integrand size = 29, antiderivative size = 168 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a^2 \csc (c+d x)}{d}-\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d} \]

[Out]

-a^2*csc(d*x+c)/d-1/16*(15*a^2+16*a*b+3*b^2)*ln(1-sin(d*x+c))/d+2*a*b*ln(sin(d*x+c))/d+1/16*(15*a^2-16*a*b+3*b
^2)*ln(1+sin(d*x+c))/d+1/8*b*sec(d*x+c)^2*(8*a+(3+7*a^2/b^2)*b*sin(d*x+c))/d+1/4*b*sec(d*x+c)^4*(2*a+(a^2+b^2)
*sin(d*x+c)/b)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 12, 1819, 1816} \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {b \sec ^4(c+d x) \left (\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}+2 a\right )}{4 d}+\frac {b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+3\right ) \sin (c+d x)+8 a\right )}{8 d}-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d} \]

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

-((a^2*Csc[c + d*x])/d) - ((15*a^2 + 16*a*b + 3*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + (2*a*b*Log[Sin[c + d*x]])
/d + ((15*a^2 - 16*a*b + 3*b^2)*Log[1 + Sin[c + d*x]])/(16*d) + (b*Sec[c + d*x]^2*(8*a + (3 + (7*a^2)/b^2)*b*S
in[c + d*x]))/(8*d) + (b*Sec[c + d*x]^4*(2*a + ((a^2 + b^2)*Sin[c + d*x])/b))/(4*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 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {b^2 (a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^7 \text {Subst}\left (\int \frac {(a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}-\frac {b^5 \text {Subst}\left (\int \frac {-4 a^2-8 a x-3 \left (1+\frac {a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac {b^3 \text {Subst}\left (\int \frac {8 a^2+16 a x+\left (3+\frac {7 a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac {b^3 \text {Subst}\left (\int \left (\frac {15 a^2+16 a b+3 b^2}{2 b^3 (b-x)}+\frac {8 a^2}{b^2 x^2}+\frac {16 a}{b^2 x}+\frac {15 a^2-16 a b+3 b^2}{2 b^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {a^2 \csc (c+d x)}{d}-\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.96 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {16 a^2 \csc (c+d x)+\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))-32 a b \log (\sin (c+d x))-\left (15 a^2-16 a b+3 b^2\right ) \log (1+\sin (c+d x))-\frac {(a+b)^2}{(-1+\sin (c+d x))^2}+\frac {(a+b) (7 a+3 b)}{-1+\sin (c+d x)}+\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(7 a-3 b) (a-b)}{1+\sin (c+d x)}}{16 d} \]

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

-1/16*(16*a^2*Csc[c + d*x] + (15*a^2 + 16*a*b + 3*b^2)*Log[1 - Sin[c + d*x]] - 32*a*b*Log[Sin[c + d*x]] - (15*
a^2 - 16*a*b + 3*b^2)*Log[1 + Sin[c + d*x]] - (a + b)^2/(-1 + Sin[c + d*x])^2 + ((a + b)*(7*a + 3*b))/(-1 + Si
n[c + d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((7*a - 3*b)*(a - b))/(1 + Sin[c + d*x]))/d

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(153\)
default	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(153\)
parallelrisch	\(\frac {-15 \left (a^{2}+\frac {16}{15} a b +\frac {1}{5} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {16}{15} a b +\frac {1}{5} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b \left (a \cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right ) a}{4}-\frac {11 b \sin \left (d x +c \right )}{8}-\frac {3 b \sin \left (3 d x +3 c \right )}{8}-\frac {7 a}{4}\right )}{2 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(261\)
norman	\(\frac {\frac {8 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2}}{2 d}-\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 b^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 b^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (11 a^{2}+5 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (11 a^{2}+5 b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (23 a^{2}+13 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (23 a^{2}+13 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {8 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (15 a^{2}-16 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (15 a^{2}+16 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(371\)
risch	\(-\frac {i \left (15 a^{2} {\mathrm e}^{9 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+40 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+8 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+16 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+18 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-22 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+48 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+40 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+8 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-48 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{i \left (d x +c \right )}-16 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}-\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}+\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(374\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(1/4/sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c
)))+2*a*b*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+b^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+
c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {32 \, a b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, a^{2} + 4 \, b^{2} + 8 \, {\left (2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(32*a*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + (15*a^2 - 16*a*b + 3*b^2)*cos(d*x + c)^4*log(
sin(d*x + c) + 1)*sin(d*x + c) - (15*a^2 + 16*a*b + 3*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c)
- 6*(5*a^2 + b^2)*cos(d*x + c)^4 + 2*(5*a^2 + b^2)*cos(d*x + c)^2 + 4*a^2 + 4*b^2 + 8*(2*a*b*cos(d*x + c)^2 +
a*b)*sin(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.97 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {32 \, a b \log \left (\sin \left (d x + c\right )\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (8 \, a b \sin \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} - 12 \, a b \sin \left (d x + c\right ) - 5 \, {\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 8 \, a^{2}\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/16*(32*a*b*log(sin(d*x + c)) + (15*a^2 - 16*a*b + 3*b^2)*log(sin(d*x + c) + 1) - (15*a^2 + 16*a*b + 3*b^2)*l
og(sin(d*x + c) - 1) - 2*(8*a*b*sin(d*x + c)^3 + 3*(5*a^2 + b^2)*sin(d*x + c)^4 - 12*a*b*sin(d*x + c) - 5*(5*a
^2 + b^2)*sin(d*x + c)^2 + 8*a^2)/(sin(d*x + c)^5 - 2*sin(d*x + c)^3 + sin(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {32 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {16 \, {\left (2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (12 \, a b \sin \left (d x + c\right )^{4} - 7 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, b^{2} \sin \left (d x + c\right )^{3} - 32 \, a b \sin \left (d x + c\right )^{2} + 9 \, a^{2} \sin \left (d x + c\right ) + 5 \, b^{2} \sin \left (d x + c\right ) + 24 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(32*a*b*log(abs(sin(d*x + c))) + (15*a^2 - 16*a*b + 3*b^2)*log(abs(sin(d*x + c) + 1)) - (15*a^2 + 16*a*b
+ 3*b^2)*log(abs(sin(d*x + c) - 1)) - 16*(2*a*b*sin(d*x + c) + a^2)/sin(d*x + c) + 2*(12*a*b*sin(d*x + c)^4 -
7*a^2*sin(d*x + c)^3 - 3*b^2*sin(d*x + c)^3 - 32*a*b*sin(d*x + c)^2 + 9*a^2*sin(d*x + c) + 5*b^2*sin(d*x + c)
+ 24*a*b)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 11.74 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {15\,a^2}{16}-a\,b+\frac {3\,b^2}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {15\,a^2}{16}+a\,b+\frac {3\,b^2}{16}\right )}{d}-\frac {a^2+{\sin \left (c+d\,x\right )}^4\,\left (\frac {15\,a^2}{8}+\frac {3\,b^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {25\,a^2}{8}+\frac {5\,b^2}{8}\right )-\frac {3\,a\,b\,\sin \left (c+d\,x\right )}{2}+a\,b\,{\sin \left (c+d\,x\right )}^3}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}+\frac {2\,a\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]

[In]

int((a + b*sin(c + d*x))^2/(cos(c + d*x)^5*sin(c + d*x)^2),x)

[Out]

(log(sin(c + d*x) + 1)*((15*a^2)/16 - a*b + (3*b^2)/16))/d - (log(sin(c + d*x) - 1)*(a*b + (15*a^2)/16 + (3*b^
2)/16))/d - (a^2 + sin(c + d*x)^4*((15*a^2)/8 + (3*b^2)/8) - sin(c + d*x)^2*((25*a^2)/8 + (5*b^2)/8) - (3*a*b*
sin(c + d*x))/2 + a*b*sin(c + d*x)^3)/(d*(sin(c + d*x) - 2*sin(c + d*x)^3 + sin(c + d*x)^5)) + (2*a*b*log(sin(
c + d*x)))/d