Integrand size = 29, antiderivative size = 249 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )^3 d}-\frac {12 a^2-a b-9 b^2}{16 (a-b) (a+b)^2 d (1-\sin (c+d x))}-\frac {\sin (c+d x)}{b d}-\frac {12 a^2+a b-9 b^2}{16 (a-b)^2 (a+b) d (1+\sin (c+d x))}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d} \] Output:
-1/16*(24*a^2+37*a*b+15*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(24*a^2-37*a* b+15*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d+a^7*ln(a+b*sin(d*x+c))/b^2/(a^2-b^2)^ 3/d-1/16*(12*a^2-a*b-9*b^2)/(a-b)/(a+b)^2/d/(1-sin(d*x+c))-sin(d*x+c)/b/d- 1/16*(12*a^2+a*b-9*b^2)/(a-b)^2/(a+b)/d/(1+sin(d*x+c))+1/4*sec(d*x+c)^4*(a -b*sin(d*x+c))/(a^2-b^2)/d
Time = 2.44 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3}+\frac {16 a^7 \log (a+b \sin (c+d x))}{(a-b)^3 b^2 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {11 a+9 b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {16 \sin (c+d x)}{b}+\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-11 a+9 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]
Output:
(-(((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((24*a^ 2 - 37*a*b + 15*b^2)*Log[1 + Sin[c + d*x]])/(a - b)^3 + (16*a^7*Log[a + b* Sin[c + d*x]])/((a - b)^3*b^2*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^ 2) + (11*a + 9*b)/((a + b)^2*(-1 + Sin[c + d*x])) - (16*Sin[c + d*x])/b + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-11*a + 9*b)/((a - b)^2*(1 + Sin[c + d *x])))/(16*d)
Time = 0.90 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3316, 27, 601, 25, 2178, 25, 2160, 2009}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^7}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b^5 \int \frac {\sin ^7(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^7 \sin ^7(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 601 |
\(\displaystyle \frac {\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\frac {a b^8}{a^2-b^2}-4 \sin ^5(c+d x) b^7-4 \sin ^3(c+d x) b^7-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) b^7}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{b^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {\frac {a b^8}{a^2-b^2}-4 \sin ^5(c+d x) b^7-4 \sin ^3(c+d x) b^7-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) b^7}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^2 d}\) |
\(\Big \downarrow \) 2178 |
\(\displaystyle \frac {\frac {\frac {\int -\frac {\frac {a \left (11 a^2-7 b^2\right ) b^8}{\left (a^2-b^2\right )^2}-8 \sin ^3(c+d x) b^7-\frac {\left (16 a^4-19 b^2 a^2+7 b^4\right ) \sin (c+d x) b^7}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {\frac {a \left (11 a^2-7 b^2\right ) b^8}{\left (a^2-b^2\right )^2}-8 \sin ^3(c+d x) b^7-\frac {\left (16 a^4-19 b^2 a^2+7 b^4\right ) \sin (c+d x) b^7}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^2 d}\) |
\(\Big \downarrow \) 2160 |
\(\displaystyle \frac {\frac {-\frac {\int \left (-\frac {8 b^4 a^7}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+8 b^4-\frac {b^6 \left (24 a^2+37 b a+15 b^2\right )}{2 (a+b)^3 (b-b \sin (c+d x))}+\frac {b^6 \left (24 a^2-37 b a+15 b^2\right )}{2 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^6 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {-\frac {b^6 \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\frac {b^6 \left (24 a^2+37 a b+15 b^2\right ) \log (b-b \sin (c+d x))}{2 (a+b)^3}+\frac {b^6 \left (24 a^2-37 a b+15 b^2\right ) \log (b \sin (c+d x)+b)}{2 (a-b)^3}-\frac {8 a^7 b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+8 b^5 \sin (c+d x)}{2 b^2}}{4 b^2}}{b^2 d}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]
Output:
((b^6*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + (-1/2*((b^6*(24*a^2 + 37*a*b + 15*b^2)*Log[b - b*Sin[c + d*x]])/(2*(a + b )^3) - (8*a^7*b^4*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^3 + (b^6*(24*a^2 - 37*a*b + 15*b^2)*Log[b + b*Sin[c + d*x]])/(2*(a - b)^3) + 8*b^5*Sin[c + d* x])/b^2 - (b^6*(4*a*(3*a^2 - 2*b^2) - b*(13*a^2 - 9*b^2)*Sin[c + d*x]))/(2 *(a^2 - b^2)^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2))/(b^2*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]
Time = 1.56 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {11 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-24 a^{2}+37 a b -15 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-11 a -9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-24 a^{2}-37 a b -15 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(203\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {11 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-24 a^{2}+37 a b -15 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-11 a -9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-24 a^{2}-37 a b -15 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(203\) |
risch | \(-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i x a}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {37 i a b c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {37 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i a^{7} c}{b^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {3 i a^{2} x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {15 i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i a^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {15 i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a^{2} c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {37 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {15 i b^{2} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i a^{2} c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {37 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i a^{7} x}{b^{2} \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {i \left (24 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-13 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+32 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+24 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+13 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{7} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {37 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {37 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(1045\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^5/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-sin(d*x+c)/b+1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(11*a-9*b)/(a-b)^2/ (1+sin(d*x+c))+1/16/(a-b)^3*(-24*a^2+37*a*b-15*b^2)*ln(1+sin(d*x+c))+1/b^2 *a^7/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/1 6*(-11*a-9*b)/(a+b)^2/(sin(d*x+c)-1)+1/16/(a+b)^3*(-24*a^2-37*a*b-15*b^2)* ln(sin(d*x+c)-1))
Time = 0.28 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.41 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + 4 \, a b^{6} - {\left (24 \, a^{5} b^{2} + 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} - 42 \, a^{2} b^{5} + 8 \, a b^{6} + 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (24 \, a^{5} b^{2} - 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} + 42 \, a^{2} b^{5} + 8 \, a b^{6} - 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (3 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + 2 \, b^{7} + 8 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (13 \, a^{4} b^{3} - 22 \, a^{2} b^{5} + 9 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/16*(16*a^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^5*b^2 - 8*a^3*b^ 4 + 4*a*b^6 - (24*a^5*b^2 + 35*a^4*b^3 - 24*a^3*b^4 - 42*a^2*b^5 + 8*a*b^6 + 15*b^7)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (24*a^5*b^2 - 35*a^4*b^3 - 24*a^3*b^4 + 42*a^2*b^5 + 8*a*b^6 - 15*b^7)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(3*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^2 - 2*(2*a^4 *b^3 - 4*a^2*b^5 + 2*b^7 + 8*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^4 - (13*a^4*b^3 - 22*a^2*b^5 + 9*b^7)*cos(d*x + c)^2)*sin(d*x + c))/ ((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^4)
\[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**5/(a+b*sin(d*x+c)),x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**5/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (13 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{3} - 6 \, a b^{2} - 4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} - \frac {16 \, \sin \left (d x + c\right )}{b}}{16 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/16*(16*a^7*log(b*sin(d*x + c) + a)/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^ 8) - (24*a^2 - 37*a*b + 15*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a *b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^ 2*b + 3*a*b^2 + b^3) - 2*((13*a^2*b - 9*b^3)*sin(d*x + c)^3 + 10*a^3 - 6*a *b^2 - 4*(3*a^3 - 2*a*b^2)*sin(d*x + c)^2 - (11*a^2*b - 7*b^3)*sin(d*x + c ))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^ 4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) - 16*sin(d*x + c)/b)/d
Time = 0.23 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{2} d - 3 \, a^{4} b^{4} d + 3 \, a^{2} b^{6} d - b^{8} d} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {\sin \left (d x + c\right )}{b d} - \frac {10 \, a^{5} - 16 \, a^{3} b^{2} + 6 \, a b^{4} + {\left (13 \, a^{4} b - 22 \, a^{2} b^{3} + 9 \, b^{5}\right )} \sin \left (d x + c\right )^{3} - 4 \, {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{4} b - 18 \, a^{2} b^{3} + 7 \, b^{5}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
a^7*log(abs(b*sin(d*x + c) + a))/(a^6*b^2*d - 3*a^4*b^4*d + 3*a^2*b^6*d - b^8*d) - 1/16*(24*a^2 + 37*a*b + 15*b^2)*log(abs(-sin(d*x + c) + 1))/(a^3* d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) - 1/16*(24*a^2 - 37*a*b + 15*b^2)*log(a bs(-sin(d*x + c) - 1))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) - sin(d*x + c)/(b*d) - 1/8*(10*a^5 - 16*a^3*b^2 + 6*a*b^4 + (13*a^4*b - 22*a^2*b^3 + 9*b^5)*sin(d*x + c)^3 - 4*(3*a^5 - 5*a^3*b^2 + 2*a*b^4)*sin(d*x + c)^2 - ( 11*a^4*b - 18*a^2*b^3 + 7*b^5)*sin(d*x + c))/((a + b)^3*(a - b)^3*d*(sin(d *x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 21.12 (sec) , antiderivative size = 627, normalized size of antiderivative = 2.52 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^5)/(a + b*sin(c + d*x)),x)
Output:
- (log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (11*b)/(8*(a - b)^2) + 3/(a - b)))/d - ((2*tan(c/2 + (d*x)/2)^4*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^2*(a*b^2 - 2*a^3))/(a^2 - b^2)^2 + (2*tan(c/2 + ( d*x)/2)^6*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^8*(a*b^ 2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^5*(24*a^4 + 9*b^ 4 - 29*a^2*b^2))/(2*b*(a^2 - b^2)^2) - (2*tan(c/2 + (d*x)/2)^3*(4*a^2 - 5* b^2))/(b*(a^2 - b^2)) - (2*tan(c/2 + (d*x)/2)^7*(4*a^2 - 5*b^2))/(b*(a^2 - b^2)) + (tan(c/2 + (d*x)/2)*(8*a^4 + 15*b^4 - 27*a^2*b^2))/(4*b*(a^4 + b^ 4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^9*(8*a^4 + 15*b^4 - 27*a^2*b^2))/(4* b*(a^4 + b^4 - 2*a^2*b^2)))/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x) /2)^2 + 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/ 2)^10 + 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(3/(a + b) - (11*b)/(8*(a + b)^ 2) + b^2/(4*(a + b)^3)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/(b^2*d) - ( a^7*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(b^8 - 3* a^2*b^6 + 3*a^4*b^4 - a^6*b^2))
Time = 0.22 (sec) , antiderivative size = 3742, normalized size of antiderivative = 15.03 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^5/(a+b*sin(d*x+c)),x)
Output:
( - 4*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**3*a**6*b + 12*cos(c + d*x )*sin(c + d*x)**4*tan(c + d*x)**3*a**4*b**3 - 12*cos(c + d*x)*sin(c + d*x) **4*tan(c + d*x)**3*a**2*b**5 + 4*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x )**3*b**7 + 38*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)*a**6*b - 114*cos( c + d*x)*sin(c + d*x)**4*tan(c + d*x)*a**4*b**3 + 114*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)*a**2*b**5 - 38*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)*b**7 + 8*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)**3*a**6*b - 24*cos (c + d*x)*sin(c + d*x)**2*tan(c + d*x)**3*a**4*b**3 + 24*cos(c + d*x)*sin( c + d*x)**2*tan(c + d*x)**3*a**2*b**5 - 8*cos(c + d*x)*sin(c + d*x)**2*tan (c + d*x)**3*b**7 - 76*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)*a**6*b + 228*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)*a**4*b**3 - 228*cos(c + d*x) *sin(c + d*x)**2*tan(c + d*x)*a**2*b**5 + 76*cos(c + d*x)*sin(c + d*x)**2* tan(c + d*x)*b**7 - 4*cos(c + d*x)*tan(c + d*x)**3*a**6*b + 12*cos(c + d*x )*tan(c + d*x)**3*a**4*b**3 - 12*cos(c + d*x)*tan(c + d*x)**3*a**2*b**5 + 4*cos(c + d*x)*tan(c + d*x)**3*b**7 + 38*cos(c + d*x)*tan(c + d*x)*a**6*b - 114*cos(c + d*x)*tan(c + d*x)*a**4*b**3 + 114*cos(c + d*x)*tan(c + d*x)* a**2*b**5 - 38*cos(c + d*x)*tan(c + d*x)*b**7 + 21*log(tan(c + d*x)**2 + 1 )*sin(c + d*x)**4*a**7 - 63*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*a**5* b**2 + 63*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*a**3*b**4 - 21*log(tan( c + d*x)**2 + 1)*sin(c + d*x)**4*a*b**6 - 42*log(tan(c + d*x)**2 + 1)*s...