Integrand size = 21, antiderivative size = 218 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {(a+b)^2 (8 a+11 b)}{16 d (1-\sin (c+d x))}-\frac {3 b \left (a^2+b^2\right ) \sin (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}-\frac {(8 a-11 b) (a-b)^2}{16 d (1+\sin (c+d x))}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \] Output:
-1/16*(a+b)*(8*a^2+37*a*b+35*b^2)*ln(1-sin(d*x+c))/d-1/16*(a-b)*(8*a^2-37* a*b+35*b^2)*ln(1+sin(d*x+c))/d-1/16*(a+b)^2*(8*a+11*b)/d/(1-sin(d*x+c))-3* b*(a^2+b^2)*sin(d*x+c)/d-3/2*a*b^2*sin(d*x+c)^2/d-1/3*b^3*sin(d*x+c)^3/d-1 /16*(8*a-11*b)*(a-b)^2/d/(1+sin(d*x+c))+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c))^ 3/d
Time = 1.16 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.91 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {3 (a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 (a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))-\frac {3 (a+b)^3}{(-1+\sin (c+d x))^2}-\frac {3 (a+b)^2 (7 a+13 b)}{-1+\sin (c+d x)}+144 b \left (a^2+b^2\right ) \sin (c+d x)+72 a b^2 \sin ^2(c+d x)+16 b^3 \sin ^3(c+d x)-\frac {3 (a-b)^3}{(1+\sin (c+d x))^2}+\frac {3 (7 a-13 b) (a-b)^2}{1+\sin (c+d x)}}{48 d} \] Input:
Integrate[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]
Output:
-1/48*(3*(a + b)*(8*a^2 + 37*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(8*a^2 - 37*a*b + 35*b^2)*Log[1 + Sin[c + d*x]] - (3*(a + b)^3)/(-1 + S in[c + d*x])^2 - (3*(a + b)^2*(7*a + 13*b))/(-1 + Sin[c + d*x]) + 144*b*(a ^2 + b^2)*Sin[c + d*x] + 72*a*b^2*Sin[c + d*x]^2 + 16*b^3*Sin[c + d*x]^3 - (3*(a - b)^3)/(1 + Sin[c + d*x])^2 + (3*(7*a - 13*b)*(a - b)^2)/(1 + Sin[ c + d*x]))/d
Time = 0.61 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3200, 531, 25, 2176, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^5 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {\frac {\int -\frac {(a+b \sin (c+d x))^2 \left (4 \sin ^4(c+d x) b^6+4 \sin ^2(c+d x) b^6+3 b^6+4 a \sin ^3(c+d x) b^5+4 a \sin (c+d x) b^5\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x))^2 \left (4 \sin ^4(c+d x) b^6+4 \sin ^2(c+d x) b^6+3 b^6+4 a \sin ^3(c+d x) b^5+4 a \sin (c+d x) b^5\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {(a+b \sin (c+d x)) \left (8 \sin ^3(c+d x) b^7+16 a \sin ^2(c+d x) b^6+21 a b^6+\left (8 a^2+27 b^2\right ) \sin (c+d x) b^5\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^4 (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {(a+b \sin (c+d x)) \left (8 \sin ^3(c+d x) b^7+16 a \sin ^2(c+d x) b^6+21 a b^6+\left (8 a^2+27 b^2\right ) \sin (c+d x) b^5\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (-8 \sin ^2(c+d x) b^6-35 b^6-24 a \sin (c+d x) b^5-24 a^2 b^4+\frac {5 \left (9 a^2+7 b^2\right ) b^6+8 a \left (a^2+9 b^2\right ) \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {5 b^5 \left (9 a^2+7 b^2\right ) \text {arctanh}(\sin (c+d x))-b^5 \left (24 a^2+35 b^2\right ) \sin (c+d x)-4 a b^4 \left (a^2+9 b^2\right ) \log \left (b^2-b^2 \sin ^2(c+d x)\right )-12 a b^6 \sin ^2(c+d x)-\frac {8}{3} b^7 \sin ^3(c+d x)}{2 b^2}}{4 b^2}}{d}\) |
Input:
Int[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]
Output:
((b^4*(a + b*Sin[c + d*x])^3)/(4*(b^2 - b^2*Sin[c + d*x]^2)^2) - ((b^4*(a + b*Sin[c + d*x])^2*(8*a + 11*b*Sin[c + d*x]))/(2*(b^2 - b^2*Sin[c + d*x]^ 2)) - (5*b^5*(9*a^2 + 7*b^2)*ArcTanh[Sin[c + d*x]] - 4*a*b^4*(a^2 + 9*b^2) *Log[b^2 - b^2*Sin[c + d*x]^2] - b^5*(24*a^2 + 35*b^2)*Sin[c + d*x] - 12*a *b^6*Sin[c + d*x]^2 - (8*b^7*Sin[c + d*x]^3)/3)/(2*b^2))/(4*b^2))/d
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
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[Pq, a + b*x^2, x], R = Coeff[PolynomialRema inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 , x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && R ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 12.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(304\) |
| default | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(304\) |
| parts | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(316\) |
| risch | \(\frac {2 i a^{3} c}{d}+\frac {18 i a \,b^{2} c}{d}-\frac {13 i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+9 i a \,b^{2} x +i a^{3} x +\frac {13 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}+\frac {3 a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (16 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+72 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+27 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+13 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+96 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+5 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+72 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-27 a^{2} b -13 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 i b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}+\frac {i b^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{d}+\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}\) | \(610\) |
Input:
int((a+b*sin(d*x+c))^3*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+3*a^2*b*(1/4*s in(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/ 8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+3*a*b^2*(1/ 4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*sin(d*x+c)^6 -3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+b^3*(1/4*sin(d*x+c)^9 /cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5/8*sin(d*x+c)^7-7/8*sin(d*x+c )^5-35/24*sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.14 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.09 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {72 \, a b^{2} \cos \left (d x + c\right )^{6} - 36 \, a b^{2} \cos \left (d x + c\right )^{4} - 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{3} + 36 \, a b^{2} - 24 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (9 \, a^{2} b + 10 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 18 \, a^{2} b + 6 \, b^{3} - 3 \, {\left (27 \, a^{2} b + 13 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^5,x, algorithm="fricas")
Output:
1/48*(72*a*b^2*cos(d*x + c)^6 - 36*a*b^2*cos(d*x + c)^4 - 3*(8*a^3 - 45*a^ 2*b + 72*a*b^2 - 35*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 12* a^3 + 36*a*b^2 - 24*(2*a^3 + 9*a*b^2)*cos(d*x + c)^2 + 2*(8*b^3*cos(d*x + c)^6 - 8*(9*a^2*b + 10*b^3)*cos(d*x + c)^4 + 18*a^2*b + 6*b^3 - 3*(27*a^2* b + 13*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*sin(d*x+c))**3*tan(d*x+c)**5,x)
Output:
Integral((a + b*sin(c + d*x))**3*tan(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right )^{3} + 72 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left ({\left (27 \, a^{2} b + 13 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 30 \, a b^{2} + 4 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (21 \, a^{2} b + 11 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/48*(16*b^3*sin(d*x + c)^3 + 72*a*b^2*sin(d*x + c)^2 + 3*(8*a^3 - 45*a^2 *b + 72*a*b^2 - 35*b^3)*log(sin(d*x + c) + 1) + 3*(8*a^3 + 45*a^2*b + 72*a *b^2 + 35*b^3)*log(sin(d*x + c) - 1) + 144*(a^2*b + b^3)*sin(d*x + c) - 6* ((27*a^2*b + 13*b^3)*sin(d*x + c)^3 - 6*a^3 - 30*a*b^2 + 4*(2*a^3 + 9*a*b^ 2)*sin(d*x + c)^2 - (21*a^2*b + 11*b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2* sin(d*x + c)^2 + 1))/d
Time = 0.52 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.12 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {2 \, b^{3} d^{2} \sin \left (d x + c\right )^{3} + 9 \, a b^{2} d^{2} \sin \left (d x + c\right )^{2} + 18 \, a^{2} b d^{2} \sin \left (d x + c\right ) + 18 \, b^{3} d^{2} \sin \left (d x + c\right )}{6 \, d^{3}} + \frac {{\left (27 \, a^{2} b + 13 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 30 \, a b^{2} + 4 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (21 \, a^{2} b + 11 \, b^{3}\right )} \sin \left (d x + c\right )}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/16*(8*a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*log(abs(sin(d*x + c) + 1))/d - 1/16*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*log(abs(sin(d*x + c) - 1))/d - 1/6*(2*b^3*d^2*sin(d*x + c)^3 + 9*a*b^2*d^2*sin(d*x + c)^2 + 18*a^2*b*d ^2*sin(d*x + c) + 18*b^3*d^2*sin(d*x + c))/d^3 + 1/8*((27*a^2*b + 13*b^3)* sin(d*x + c)^3 - 6*a^3 - 30*a*b^2 + 4*(2*a^3 + 9*a*b^2)*sin(d*x + c)^2 - ( 21*a^2*b + 11*b^3)*sin(d*x + c))/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1 )^2)
Time = 20.17 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.35 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^5*(a + b*sin(c + d*x))^3,x)
Output:
(log(tan(c/2 + (d*x)/2)^2 + 1)*(9*a*b^2 + a^3))/d - (tan(c/2 + (d*x)/2)^4* (18*a*b^2 + 2*a^3) - tan(c/2 + (d*x)/2)^2*(18*a*b^2 + 2*a^3) - tan(c/2 + ( d*x)/2)*((45*a^2*b)/4 + (35*b^3)/4) + tan(c/2 + (d*x)/2)^10*(18*a*b^2 + 2* a^3) - tan(c/2 + (d*x)/2)^12*(18*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^6*(48 *a*b^2 + 16*a^3) + tan(c/2 + (d*x)/2)^8*(48*a*b^2 + 16*a^3) + tan(c/2 + (d *x)/2)^7*(33*a^2*b - 17*b^3) + tan(c/2 + (d*x)/2)^3*((15*a^2*b)/2 + (35*b^ 3)/6) + tan(c/2 + (d*x)/2)^11*((15*a^2*b)/2 + (35*b^3)/6) - tan(c/2 + (d*x )/2)^13*((45*a^2*b)/4 + (35*b^3)/4) + tan(c/2 + (d*x)/2)^5*((141*a^2*b)/4 + (329*b^3)/12) + tan(c/2 + (d*x)/2)^9*((141*a^2*b)/4 + (329*b^3)/12))/(d* (tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + 3*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 - 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(a - b)*(8*a^2 - 37*a*b + 35*b^2))/(8*d) - (log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(37*a*b + 8*a^2 + 35*b^2))/(8*d)
Time = 0.18 (sec) , antiderivative size = 824, normalized size of antiderivative = 3.78 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*sin(d*x+c))^3*tan(d*x+c)^5,x)
Output:
(12*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*a**3 - 24*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2*a**3 + 12*log(tan(c + d*x)**2 + 1)*a**3 + 216*log(ta n((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a*b**2 - 432*log(tan((c + d*x)/2)** 2 + 1)*sin(c + d*x)**2*a*b**2 + 216*log(tan((c + d*x)/2)**2 + 1)*a*b**2 - 135*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b - 216*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**4*a*b**2 - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**3 + 270*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + 432* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 + 210*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**2*b**3 - 135*log(tan((c + d*x)/2) - 1)*a**2*b - 216*l og(tan((c + d*x)/2) - 1)*a*b**2 - 105*log(tan((c + d*x)/2) - 1)*b**3 + 135 *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b - 216*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**4*a*b**2 + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x )**4*b**3 - 270*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b + 432*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 - 210*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 + 135*log(tan((c + d*x)/2) + 1)*a**2*b - 216*log( tan((c + d*x)/2) + 1)*a*b**2 + 105*log(tan((c + d*x)/2) + 1)*b**3 - 8*sin( c + d*x)**7*b**3 - 36*sin(c + d*x)**6*a*b**2 - 72*sin(c + d*x)**5*a**2*b - 56*sin(c + d*x)**5*b**3 + 6*sin(c + d*x)**4*tan(c + d*x)**4*a**3 - 12*sin (c + d*x)**4*tan(c + d*x)**2*a**3 + 162*sin(c + d*x)**4*a*b**2 + 225*sin(c + d*x)**3*a**2*b + 175*sin(c + d*x)**3*b**3 - 12*sin(c + d*x)**2*tan(c...