\(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1500]

Optimal result

Integrand size = 29, antiderivative size = 185 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 a b \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {\left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d} \]

[Out]

-2*a*b*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d-1/8*(12*a^2+15*a*b+4*b^2)*ln(1-sin(d*x+c))/d+(3*a^2+b^2)*ln(sin(d*x
+c))/d-1/8*(12*a^2-15*a*b+4*b^2)*ln(1+sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a^2+b^2+2*a*b*sin(d*x+c))/d+1/4*sec(d*x+
c)^2*(4*a^2+2*b^2+7*a*b*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 185, 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} \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {\left (12 a^2-15 a b+4 b^2\right ) \log (\sin (c+d x)+1)}{8 d}+\frac {\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a b \csc (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(-2*a*b*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - ((12*a^2 + 15*a*b + 4*b^2)*Log[1 - Sin[c + d*x]])/(8*d)
+ ((3*a^2 + b^2)*Log[Sin[c + d*x]])/d - ((12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*
x]^4*(a^2 + b^2 + 2*a*b*Sin[c + d*x]))/(4*d) + (Sec[c + d*x]^2*(2*(2*a^2 + b^2) + 7*a*b*Sin[c + d*x]))/(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^3 (a+x)^2}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^8 \text {Subst}\left (\int \frac {(a+x)^2}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {b^6 \text {Subst}\left (\int \frac {-4 a^2-8 a x-4 \left (1+\frac {a^2}{b^2}\right ) x^2-\frac {6 a x^3}{b^2}}{x^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}+\frac {b^4 \text {Subst}\left (\int \frac {8 a^2+16 a x+8 \left (1+\frac {2 a^2}{b^2}\right ) x^2+\frac {14 a x^3}{b^2}}{x^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}+\frac {b^4 \text {Subst}\left (\int \left (\frac {12 a^2+15 a b+4 b^2}{b^4 (b-x)}+\frac {8 a^2}{b^2 x^3}+\frac {16 a}{b^2 x^2}+\frac {8 \left (3 a^2+b^2\right )}{b^4 x}+\frac {-12 a^2+15 a b-4 b^2}{b^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {2 a b \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {\left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.69 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.98 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-32 a b \csc (c+d x)-8 a^2 \csc ^2(c+d x)-2 \left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))+16 \left (3 a^2+b^2\right ) \log (\sin (c+d x))-2 \left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}-\frac {(a+b) (9 a+5 b)}{-1+\sin (c+d x)}+\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(9 a-5 b) (a-b)}{1+\sin (c+d x)}}{16 d} \]

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(-32*a*b*Csc[c + d*x] - 8*a^2*Csc[c + d*x]^2 - 2*(12*a^2 + 15*a*b + 4*b^2)*Log[1 - Sin[c + d*x]] + 16*(3*a^2 +
b^2)*Log[Sin[c + d*x]] - 2*(12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2
- ((a + b)*(9*a + 5*b))/(-1 + Sin[c + d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((9*a - 5*b)*(a - b))/(1 + Sin[
c + d*x]))/(16*d)

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(165\)
default	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(165\)
parallelrisch	\(\frac {-384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {5}{4} a b +\frac {1}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {5}{4} a b +\frac {1}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-27 \left (\cos \left (2 d x +2 c \right )+\frac {2 \cos \left (4 d x +4 c \right )}{9}-\frac {\cos \left (6 d x +6 c \right )}{9}+\frac {2}{27}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-160 b \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-32 b^{2} \left (\frac {3 \cos \left (4 d x +4 c \right )}{4}-\frac {7}{4}+\cos \left (2 d x +2 c \right )\right )}{32 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(303\)
risch	\(-\frac {i \left (12 i a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+4 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+15 a b \,{\mathrm e}^{11 i \left (d x +c \right )}+24 i a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+4 i b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+25 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+24 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-22 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-8 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+22 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-25 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-15 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\)	\(426\)

[In]

int(csc(d*x+c)^3*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)^2/cos(d*x+c)^4+3/4/sin(d*x+c)^2/cos(d*x+c)^2-3/2/sin(d*x+c)^2+3*ln(tan(d*x+c)))+2*a*b
*(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)))+b^2*
(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.54 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} + 8 \, {\left ({\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left ({\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, a b \cos \left (d x + c\right )^{4} - 5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, a b \sin \left (d x + c\right )^{5} - 25 \, a b \sin \left (d x + c\right )^{3} + 2 \, {\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a b \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{8 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(sin(d*x + c) + 1) + (12*a^2 + 15*a*b + 4*b^2)*log(sin(d*x + c) - 1) - 8*(3
*a^2 + b^2)*log(sin(d*x + c)) + 2*(15*a*b*sin(d*x + c)^5 - 25*a*b*sin(d*x + c)^3 + 2*(3*a^2 + b^2)*sin(d*x + c
)^4 + 8*a*b*sin(d*x + c) - 3*(3*a^2 + b^2)*sin(d*x + c)^2 + 2*a^2)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*
x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.03 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 8 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (15 \, a b \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{4} + 2 \, b^{2} \sin \left (d x + c\right )^{4} - 25 \, a b \sin \left (d x + c\right )^{3} - 9 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 8 \, a b \sin \left (d x + c\right ) + 2 \, a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{8 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(abs(sin(d*x + c) + 1)) + (12*a^2 + 15*a*b + 4*b^2)*log(abs(sin(d*x + c) -
1)) - 8*(3*a^2 + b^2)*log(abs(sin(d*x + c))) + 2*(15*a*b*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^4 + 2*b^2*sin(d*x
+ c)^4 - 25*a*b*sin(d*x + c)^3 - 9*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^2 + 8*a*b*sin(d*x + c) + 2*a^2)/(s
in(d*x + c)^3 - sin(d*x + c))^2)/d

Mupad [B] (verification not implemented)

Time = 12.45 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.05 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (3\,a^2+b^2\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a^2}{2}-\frac {15\,a\,b}{8}+\frac {b^2}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a^2}{2}+\frac {15\,a\,b}{8}+\frac {b^2}{2}\right )}{d}-\frac {\frac {a^2}{2}+{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^2}{2}+\frac {b^2}{2}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {9\,a^2}{4}+\frac {3\,b^2}{4}\right )+2\,a\,b\,\sin \left (c+d\,x\right )-\frac {25\,a\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {15\,a\,b\,{\sin \left (c+d\,x\right )}^5}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )} \]

[In]

int((a + b*sin(c + d*x))^2/(cos(c + d*x)^5*sin(c + d*x)^3),x)

[Out]

(log(sin(c + d*x))*(3*a^2 + b^2))/d - (log(sin(c + d*x) + 1)*((3*a^2)/2 - (15*a*b)/8 + b^2/2))/d - (log(sin(c
+ d*x) - 1)*((15*a*b)/8 + (3*a^2)/2 + b^2/2))/d - (a^2/2 + sin(c + d*x)^4*((3*a^2)/2 + b^2/2) - sin(c + d*x)^2
*((9*a^2)/4 + (3*b^2)/4) + 2*a*b*sin(c + d*x) - (25*a*b*sin(c + d*x)^3)/4 + (15*a*b*sin(c + d*x)^5)/4)/(d*(sin
(c + d*x)^2 - 2*sin(c + d*x)^4 + sin(c + d*x)^6))