Integrand size = 21, antiderivative size = 160 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {\left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac {\left (15 a b-4 \left (a^2+3 b^2\right )\right ) \log (1+\sin (c+d x))}{8 d}-\frac {2 a b \sin (c+d x)}{d}-\frac {b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) \left (4 a^2+5 b^2+9 a b \sin (c+d x)\right )}{4 d} \] Output:
-1/8*(4*a^2+15*a*b+12*b^2)*ln(1-sin(d*x+c))/d+1/8*(-4*a^2+15*a*b-12*b^2)*l n(1+sin(d*x+c))/d-2*a*b*sin(d*x+c)/d-1/2*b^2*sin(d*x+c)^2/d+1/4*sec(d*x+c) ^4*(a+b*sin(d*x+c))^2/d-1/4*sec(d*x+c)^2*(4*a^2+5*b^2+9*a*b*sin(d*x+c))/d
Time = 2.54 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {-2 \left (4 a^2+15 a b+12 b^2\right ) \log (1-\sin (c+d x))-2 \left (4 a^2-15 a b+12 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}+\frac {(a+b) (7 a+11 b)}{-1+\sin (c+d x)}-32 a b \sin (c+d x)-8 b^2 \sin ^2(c+d x)+\frac {(a-b)^2}{(1+\sin (c+d x))^2}-\frac {(7 a-11 b) (a-b)}{1+\sin (c+d x)}}{16 d} \] Input:
Integrate[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]
Output:
(-2*(4*a^2 + 15*a*b + 12*b^2)*Log[1 - Sin[c + d*x]] - 2*(4*a^2 - 15*a*b + 12*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2 + ((a + b) *(7*a + 11*b))/(-1 + Sin[c + d*x]) - 32*a*b*Sin[c + d*x] - 8*b^2*Sin[c + d *x]^2 + (a - b)^2/(1 + Sin[c + d*x])^2 - ((7*a - 11*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)
Time = 0.51 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11, 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, 27, 2176, 25, 2341, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^5 (a+b \sin (c+d x))^2dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {\frac {\int -\frac {2 (a+b \sin (c+d x)) \left (2 \sin ^4(c+d x) b^6+2 \sin ^2(c+d x) b^6+b^6+2 a \sin ^3(c+d x) b^5+2 a \sin (c+d x) b^5\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x)) \left (2 \sin ^4(c+d x) b^6+2 \sin ^2(c+d x) b^6+b^6+2 a \sin ^3(c+d x) b^5+2 a \sin (c+d x) b^5\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {4 \sin ^3(c+d x) b^7+8 a \sin ^2(c+d x) b^6+7 a b^6+4 \left (a^2+2 b^2\right ) \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^4 (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {4 \sin ^3(c+d x) b^7+8 a \sin ^2(c+d x) b^6+7 a b^6+4 \left (a^2+2 b^2\right ) \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (-4 \sin (c+d x) b^5-8 a b^4+\frac {15 a b^6+4 \left (a^2+3 b^2\right ) \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}\right )d(b \sin (c+d x))}{2 b^2}}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) (4 a+5 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {-2 b^4 \left (a^2+3 b^2\right ) \log \left (b^2-b^2 \sin ^2(c+d x)\right )+15 a b^5 \text {arctanh}(\sin (c+d x))-8 a b^5 \sin (c+d x)-2 b^6 \sin ^2(c+d x)}{2 b^2}}{2 b^2}}{d}\) |
Input:
Int[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]
Output:
((b^4*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2)^2) - ((b^4*(a + b*Sin[c + d*x])*(4*a + 5*b*Sin[c + d*x]))/(2*(b^2 - b^2*Sin[c + d*x]^2)) - (15*a*b^5*ArcTanh[Sin[c + d*x]] - 2*b^4*(a^2 + 3*b^2)*Log[b^2 - b^2*Sin [c + d*x]^2] - 8*a*b^5*Sin[c + d*x] - 2*b^6*Sin[c + d*x]^2)/(2*b^2))/(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_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] : > 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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 4.11 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.28
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(205\) |
| default | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(205\) |
| parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(214\) |
| risch | \(-\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+3 i b^{2} x +\frac {{\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+i a^{2} x +\frac {{\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {i \left (8 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+12 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+9 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+16 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+a b \,{\mathrm e}^{5 i \left (d x +c \right )}+8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {6 i b^{2} c}{d}+\frac {2 i a^{2} c}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}\) | \(389\) |
Input:
int((a+b*sin(d*x+c))^2*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+2*a*b*(1/4*sin (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)))+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*s in(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c))))
Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {4 \, b^{2} \cos \left (d x + c\right )^{6} - 2 \, b^{2} \cos \left (d x + c\right )^{4} - {\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} - 2 \, {\left (8 \, a b \cos \left (d x + c\right )^{4} + 9 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="fricas")
Output:
1/8*(4*b^2*cos(d*x + c)^6 - 2*b^2*cos(d*x + c)^4 - (4*a^2 - 15*a*b + 12*b^ 2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (4*a^2 + 15*a*b + 12*b^2)*cos(d* x + c)^4*log(-sin(d*x + c) + 1) - 4*(2*a^2 + 3*b^2)*cos(d*x + c)^2 + 2*a^2 + 2*b^2 - 2*(8*a*b*cos(d*x + c)^4 + 9*a*b*cos(d*x + c)^2 - 2*a*b)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*sin(d*x+c))**2*tan(d*x+c)**5,x)
Output:
Integral((a + b*sin(c + d*x))**2*tan(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {4 \, b^{2} \sin \left (d x + c\right )^{2} + 16 \, a b \sin \left (d x + c\right ) + {\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (9 \, a b \sin \left (d x + c\right )^{3} - 7 \, a b \sin \left (d x + c\right ) + 2 \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \] Input:
integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/8*(4*b^2*sin(d*x + c)^2 + 16*a*b*sin(d*x + c) + (4*a^2 - 15*a*b + 12*b^ 2)*log(sin(d*x + c) + 1) + (4*a^2 + 15*a*b + 12*b^2)*log(sin(d*x + c) - 1) - 2*(9*a*b*sin(d*x + c)^3 - 7*a*b*sin(d*x + c) + 2*(2*a^2 + 3*b^2)*sin(d* x + c)^2 - 3*a^2 - 5*b^2)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.48 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {{\left (4 \, a^{2} - 15 \, a b + 12 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{8 \, d} - \frac {{\left (4 \, a^{2} + 15 \, a b + 12 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{8 \, d} - \frac {b^{2} d \sin \left (d x + c\right )^{2} + 4 \, a b d \sin \left (d x + c\right )}{2 \, d^{2}} + \frac {9 \, a b \sin \left (d x + c\right )^{3} - 7 \, a b \sin \left (d x + c\right ) + 2 \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, b^{2}}{4 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/8*(4*a^2 - 15*a*b + 12*b^2)*log(abs(sin(d*x + c) + 1))/d - 1/8*(4*a^2 + 15*a*b + 12*b^2)*log(abs(sin(d*x + c) - 1))/d - 1/2*(b^2*d*sin(d*x + c)^2 + 4*a*b*d*sin(d*x + c))/d^2 + 1/4*(9*a*b*sin(d*x + c)^3 - 7*a*b*sin(d*x + c) + 2*(2*a^2 + 3*b^2)*sin(d*x + c)^2 - 3*a^2 - 5*b^2)/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 20.34 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.36 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+3\,b^2\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a^2-\frac {15\,a\,b}{4}+3\,b^2\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a^2+\frac {15\,a\,b}{4}+3\,b^2\right )}{d}-\frac {-\frac {15\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{2}+\left (-2\,a^2-6\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {25\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\left (4\,a^2+12\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,a^2+4\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,a^2+12\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {25\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\left (-2\,a^2-6\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {15\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(c + d*x)^5*(a + b*sin(c + d*x))^2,x)
Output:
(log(tan(c/2 + (d*x)/2)^2 + 1)*(a^2 + 3*b^2))/d - (log(tan(c/2 + (d*x)/2) + 1)*(a^2 - (15*a*b)/4 + 3*b^2))/d - (log(tan(c/2 + (d*x)/2) - 1)*((15*a*b )/4 + a^2 + 3*b^2))/d - (tan(c/2 + (d*x)/2)^4*(4*a^2 + 12*b^2) - tan(c/2 + (d*x)/2)^10*(2*a^2 + 6*b^2) - tan(c/2 + (d*x)/2)^2*(2*a^2 + 6*b^2) + tan( c/2 + (d*x)/2)^6*(12*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^8*(4*a^2 + 12*b^2) + (25*a*b*tan(c/2 + (d*x)/2)^3)/2 + 11*a*b*tan(c/2 + (d*x)/2)^5 + 11*a*b*t an(c/2 + (d*x)/2)^7 + (25*a*b*tan(c/2 + (d*x)/2)^9)/2 - (15*a*b*tan(c/2 + (d*x)/2)^11)/2 - (15*a*b*tan(c/2 + (d*x)/2))/2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^12 - 1))
Time = 0.18 (sec) , antiderivative size = 608, normalized size of antiderivative = 3.80 \[ \int (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*sin(d*x+c))^2*tan(d*x+c)^5,x)
Output:
(2*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*a**2 - 4*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2*a**2 + 2*log(tan(c + d*x)**2 + 1)*a**2 + 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*b**2 - 24*log(tan((c + d*x)/2)**2 + 1)* sin(c + d*x)**2*b**2 + 12*log(tan((c + d*x)/2)**2 + 1)*b**2 - 15*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**4*a*b - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**2 + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b + 24*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 - 15*log(tan((c + d*x)/2) - 1) *a*b - 12*log(tan((c + d*x)/2) - 1)*b**2 + 15*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**4*a*b - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 - 30 *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b + 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 + 15*log(tan((c + d*x)/2) + 1)*a*b - 12*log(tan(( c + d*x)/2) + 1)*b**2 - 2*sin(c + d*x)**6*b**2 - 8*sin(c + d*x)**5*a*b + s in(c + d*x)**4*tan(c + d*x)**4*a**2 - 2*sin(c + d*x)**4*tan(c + d*x)**2*a* *2 + 9*sin(c + d*x)**4*b**2 + 25*sin(c + d*x)**3*a*b - 2*sin(c + d*x)**2*t an(c + d*x)**4*a**2 + 4*sin(c + d*x)**2*tan(c + d*x)**2*a**2 - 6*sin(c + d *x)**2*b**2 - 15*sin(c + d*x)*a*b + tan(c + d*x)**4*a**2 - 2*tan(c + d*x)* *2*a**2)/(4*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))