Integrand size = 28, antiderivative size = 161 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {e f x}{2 a d}+\frac {f^2 x^2}{4 a d}+\frac {2 f (e+f x) \cos (c+d x)}{a d^2}-\frac {2 f^2 \sin (c+d x)}{a d^3}+\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a d^2}+\frac {f^2 \sin ^2(c+d x)}{4 a d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 a d} \]
1/2*e*f*x/a/d+1/4*f^2*x^2/a/d+2*f*(f*x+e)*cos(d*x+c)/a/d^2-2*f^2*sin(d*x+c )/a/d^3+(f*x+e)^2*sin(d*x+c)/a/d-1/2*f*(f*x+e)*cos(d*x+c)*sin(d*x+c)/a/d^2 +1/4*f^2*sin(d*x+c)^2/a/d^3-1/2*(f*x+e)^2*sin(d*x+c)^2/a/d
Time = 0.95 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.59 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {16 d f (e+f x) \cos (c+d x)+\left (-f^2+2 d^2 (e+f x)^2\right ) \cos (2 (c+d x))-4 \left (-2 \left (-2 f^2+d^2 (e+f x)^2\right )+d f (e+f x) \cos (c+d x)\right ) \sin (c+d x)}{8 a d^3} \]
(16*d*f*(e + f*x)*Cos[c + d*x] + (-f^2 + 2*d^2*(e + f*x)^2)*Cos[2*(c + d*x )] - 4*(-2*(-2*f^2 + d^2*(e + f*x)^2) + d*f*(e + f*x)*Cos[c + d*x])*Sin[c + d*x])/(8*a*d^3)
Time = 0.75 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.91, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5034, 3042, 3777, 25, 3042, 3777, 3042, 3117, 4904, 3042, 3791, 17}
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.
\(\displaystyle \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 5034 |
\(\displaystyle \frac {\int (e+f x)^2 \cos (c+d x)dx}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x)^2 \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {\frac {2 f \int -((e+f x) \sin (c+d x))dx}{d}+\frac {(e+f x)^2 \sin (c+d x)}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \int (e+f x) \sin (c+d x)dx}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \int (e+f x) \sin (c+d x)dx}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \int \cos (c+d x)dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 4904 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\frac {(e+f x)^2 \sin ^2(c+d x)}{2 d}-\frac {f \int (e+f x) \sin ^2(c+d x)dx}{d}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\frac {(e+f x)^2 \sin ^2(c+d x)}{2 d}-\frac {f \int (e+f x) \sin (c+d x)^2dx}{d}}{a}\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\frac {(e+f x)^2 \sin ^2(c+d x)}{2 d}-\frac {f \left (\frac {1}{2} \int (e+f x)dx+\frac {f \sin ^2(c+d x)}{4 d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 d}\right )}{d}}{a}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}}{a}-\frac {\frac {(e+f x)^2 \sin ^2(c+d x)}{2 d}-\frac {f \left (\frac {f \sin ^2(c+d x)}{4 d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^2}{4 f}\right )}{d}}{a}\) |
(((e + f*x)^2*Sin[c + d*x])/d - (2*f*(-(((e + f*x)*Cos[c + d*x])/d) + (f*S in[c + d*x])/d^2))/d)/a - (((e + f*x)^2*Sin[c + d*x]^2)/(2*d) - (f*((e + f *x)^2/(4*f) - ((e + f*x)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (f*Sin[c + d*x ]^2)/(4*d^2)))/d)/a
3.3.64.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x _)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) , x] - Simp[d*(m/(b*(n + 1))) Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.) *Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Simp[1/b Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*Sin[ c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^ 2 - b^2, 0]
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(\frac {\left (2 \left (f x +e \right )^{2} d^{2}-f^{2}\right ) \cos \left (2 d x +2 c \right )-2 d f \left (f x +e \right ) \sin \left (2 d x +2 c \right )+8 \left (\left (f x +e \right )^{2} d^{2}-2 f^{2}\right ) \sin \left (d x +c \right )+16 d f \left (f x +e \right ) \cos \left (d x +c \right )-2 d^{2} e^{2}+16 d e f +f^{2}}{8 a \,d^{3}}\) | \(112\) |
risch | \(\frac {2 f \left (f x +e \right ) \cos \left (d x +c \right )}{a \,d^{2}}+\frac {\left (d^{2} x^{2} f^{2}+2 f e x \,d^{2}+d^{2} e^{2}-2 f^{2}\right ) \sin \left (d x +c \right )}{a \,d^{3}}+\frac {\left (2 d^{2} x^{2} f^{2}+4 f e x \,d^{2}+2 d^{2} e^{2}-f^{2}\right ) \cos \left (2 d x +2 c \right )}{8 a \,d^{3}}-\frac {f \left (f x +e \right ) \sin \left (2 d x +2 c \right )}{4 d^{2} a}\) | \(139\) |
derivativedivides | \(-\frac {-\frac {c^{2} f^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+c d e f \left (\cos ^{2}\left (d x +c \right )\right )-2 c \,f^{2} \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )-\frac {d^{2} e^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+2 d e f \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )+f^{2} \left (-\frac {\left (d x +c \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\left (d x +c \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (d x +c \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{4}\right )-\sin \left (d x +c \right ) c^{2} f^{2}+2 \sin \left (d x +c \right ) c d e f +2 c \,f^{2} \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-\sin \left (d x +c \right ) d^{2} e^{2}-2 d e f \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-f^{2} \left (\left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{3} a}\) | \(339\) |
default | \(-\frac {-\frac {c^{2} f^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+c d e f \left (\cos ^{2}\left (d x +c \right )\right )-2 c \,f^{2} \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )-\frac {d^{2} e^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+2 d e f \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )+f^{2} \left (-\frac {\left (d x +c \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\left (d x +c \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (d x +c \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{4}\right )-\sin \left (d x +c \right ) c^{2} f^{2}+2 \sin \left (d x +c \right ) c d e f +2 c \,f^{2} \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-\sin \left (d x +c \right ) d^{2} e^{2}-2 d e f \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-f^{2} \left (\left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{3} a}\) | \(339\) |
norman | \(\frac {\frac {4 e f}{d^{2} a}+\frac {\left (5 d e f -3 f^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{3}}+\frac {\left (7 d e f -3 f^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{3}}+\frac {\left (2 d^{2} e^{2}+d e f -4 f^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{3}}+\frac {\left (2 d^{2} e^{2}+3 d e f -4 f^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,d^{3}}+\frac {\left (2 d^{2} e^{2}+4 d e f -7 f^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{3}}+\frac {\left (2 d^{2} e^{2}+8 d e f -7 f^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{3}}+\frac {f^{2} x^{2}}{4 a d}+\frac {f \left (d e +4 f \right ) x}{2 a \,d^{2}}+\frac {9 f^{2} x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {3 f^{2} x^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {11 f^{2} x^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {11 f^{2} x^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {3 f^{2} x^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {9 f^{2} x^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {f^{2} x^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {f \left (d e -4 f \right ) x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {f \left (3 d e -2 f \right ) x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {f \left (3 d e +2 f \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {f \left (9 d e -2 f \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {f \left (9 d e +2 f \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a \,d^{2}}+\frac {f \left (11 d e -4 f \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {f \left (11 d e +4 f \right ) x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(648\) |
1/8*((2*(f*x+e)^2*d^2-f^2)*cos(2*d*x+2*c)-2*d*f*(f*x+e)*sin(2*d*x+2*c)+8*( (f*x+e)^2*d^2-2*f^2)*sin(d*x+c)+16*d*f*(f*x+e)*cos(d*x+c)-2*d^2*e^2+16*d*e *f+f^2)/a/d^3
Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.93 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {d^{2} f^{2} x^{2} + 2 \, d^{2} e f x - {\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - f^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \, {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right ) - 2 \, {\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - 4 \, f^{2} - {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, a d^{3}} \]
-1/4*(d^2*f^2*x^2 + 2*d^2*e*f*x - (2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 - f^2)*cos(d*x + c)^2 - 8*(d*f^2*x + d*e*f)*cos(d*x + c) - 2*(2*d^2*f^2*x ^2 + 4*d^2*e*f*x + 2*d^2*e^2 - 4*f^2 - (d*f^2*x + d*e*f)*cos(d*x + c))*sin (d*x + c))/(a*d^3)
Leaf count of result is larger than twice the leaf count of optimal. 1528 vs. \(2 (141) = 282\).
Time = 3.31 (sec) , antiderivative size = 1528, normalized size of antiderivative = 9.49 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((8*d**2*e**2*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 8*d**2*e**2*tan(c/2 + d*x/2)** 2/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8*d**2*e**2*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*t an(c/2 + d*x/2)**2 + 4*a*d**3) + 2*d**2*e*f*x*tan(c/2 + d*x/2)**4/(4*a*d** 3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*d**2 *e*f*x*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/ 2 + d*x/2)**2 + 4*a*d**3) - 12*d**2*e*f*x*tan(c/2 + d*x/2)**2/(4*a*d**3*ta n(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*d**2*e*f *x*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x /2)**2 + 4*a*d**3) + 2*d**2*e*f*x/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3 *tan(c/2 + d*x/2)**2 + 4*a*d**3) + d**2*f**2*x**2*tan(c/2 + d*x/2)**4/(4*a *d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8*d **2*f**2*x**2*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3 *tan(c/2 + d*x/2)**2 + 4*a*d**3) - 6*d**2*f**2*x**2*tan(c/2 + d*x/2)**2/(4 *a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8 *d**2*f**2*x**2*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3* tan(c/2 + d*x/2)**2 + 4*a*d**3) + d**2*f**2*x**2/(4*a*d**3*tan(c/2 + d*x/2 )**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 4*d*e*f*tan(c/2 + d*x/2) **3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*...
Time = 0.21 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.80 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} e^{2}}{a} - \frac {8 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c e f}{a d} + \frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c^{2} f^{2}}{a d^{2}} - \frac {2 \, {\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} e f}{a d} + \frac {2 \, {\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} c f^{2}}{a d^{2}} - \frac {{\left ({\left (2 \, {\left (d x + c\right )}^{2} - 1\right )} \cos \left (2 \, d x + 2 \, c\right ) + 16 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (2 \, d x + 2 \, c\right ) + 8 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} f^{2}}{a d^{2}}}{8 \, d} \]
-1/8*(4*(sin(d*x + c)^2 - 2*sin(d*x + c))*e^2/a - 8*(sin(d*x + c)^2 - 2*si n(d*x + c))*c*e*f/(a*d) + 4*(sin(d*x + c)^2 - 2*sin(d*x + c))*c^2*f^2/(a*d ^2) - 2*(2*(d*x + c)*cos(2*d*x + 2*c) + 8*(d*x + c)*sin(d*x + c) + 8*cos(d *x + c) - sin(2*d*x + 2*c))*e*f/(a*d) + 2*(2*(d*x + c)*cos(2*d*x + 2*c) + 8*(d*x + c)*sin(d*x + c) + 8*cos(d*x + c) - sin(2*d*x + 2*c))*c*f^2/(a*d^2 ) - ((2*(d*x + c)^2 - 1)*cos(2*d*x + 2*c) + 16*(d*x + c)*cos(d*x + c) - 2* (d*x + c)*sin(2*d*x + 2*c) + 8*((d*x + c)^2 - 2)*sin(d*x + c))*f^2/(a*d^2) )/d
Leaf count of result is larger than twice the leaf count of optimal. 2190 vs. \(2 (151) = 302\).
Time = 0.42 (sec) , antiderivative size = 2190, normalized size of antiderivative = 13.60 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
1/8*(2*d^2*f^2*x^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 16*d^2*f^2*x^2*tan(1/2*d* x)^4*tan(1/2*c)^3 - 16*d^2*f^2*x^2*tan(1/2*d*x)^3*tan(1/2*c)^4 + 4*d^2*e*f *x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 12*d^2*f^2*x^2*tan(1/2*d*x)^4*tan(1/2*c)^ 2 - 32*d^2*f^2*x^2*tan(1/2*d*x)^3*tan(1/2*c)^3 - 32*d^2*e*f*x*tan(1/2*d*x) ^4*tan(1/2*c)^3 - 12*d^2*f^2*x^2*tan(1/2*d*x)^2*tan(1/2*c)^4 - 32*d^2*e*f* x*tan(1/2*d*x)^3*tan(1/2*c)^4 + 2*d^2*e^2*tan(1/2*d*x)^4*tan(1/2*c)^4 + 16 *d*f^2*x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 16*d^2*f^2*x^2*tan(1/2*d*x)^4*tan(1 /2*c) - 24*d^2*e*f*x*tan(1/2*d*x)^4*tan(1/2*c)^2 - 64*d^2*e*f*x*tan(1/2*d* x)^3*tan(1/2*c)^3 - 16*d^2*e^2*tan(1/2*d*x)^4*tan(1/2*c)^3 + 8*d*f^2*x*tan (1/2*d*x)^4*tan(1/2*c)^3 - 16*d^2*f^2*x^2*tan(1/2*d*x)*tan(1/2*c)^4 - 24*d ^2*e*f*x*tan(1/2*d*x)^2*tan(1/2*c)^4 - 16*d^2*e^2*tan(1/2*d*x)^3*tan(1/2*c )^4 + 8*d*f^2*x*tan(1/2*d*x)^3*tan(1/2*c)^4 + 16*d*e*f*tan(1/2*d*x)^4*tan( 1/2*c)^4 + 2*d^2*f^2*x^2*tan(1/2*d*x)^4 + 32*d^2*f^2*x^2*tan(1/2*d*x)^3*ta n(1/2*c) - 32*d^2*e*f*x*tan(1/2*d*x)^4*tan(1/2*c) + 72*d^2*f^2*x^2*tan(1/2 *d*x)^2*tan(1/2*c)^2 - 12*d^2*e^2*tan(1/2*d*x)^4*tan(1/2*c)^2 + 32*d^2*f^2 *x^2*tan(1/2*d*x)*tan(1/2*c)^3 - 32*d^2*e^2*tan(1/2*d*x)^3*tan(1/2*c)^3 - 64*d*f^2*x*tan(1/2*d*x)^3*tan(1/2*c)^3 + 8*d*e*f*tan(1/2*d*x)^4*tan(1/2*c) ^3 + 2*d^2*f^2*x^2*tan(1/2*c)^4 - 32*d^2*e*f*x*tan(1/2*d*x)*tan(1/2*c)^4 - 12*d^2*e^2*tan(1/2*d*x)^2*tan(1/2*c)^4 + 8*d*e*f*tan(1/2*d*x)^3*tan(1/2*c )^4 - f^2*tan(1/2*d*x)^4*tan(1/2*c)^4 + 16*d^2*f^2*x^2*tan(1/2*d*x)^3 +...
Time = 2.89 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.16 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {8\,d^2\,e^2\,\sin \left (c+d\,x\right )-f^2\,\cos \left (2\,c+2\,d\,x\right )-16\,f^2\,\sin \left (c+d\,x\right )+2\,d^2\,e^2\,\cos \left (2\,c+2\,d\,x\right )+8\,d^2\,f^2\,x^2\,\sin \left (c+d\,x\right )-2\,d\,e\,f\,\sin \left (2\,c+2\,d\,x\right )+16\,d\,f^2\,x\,\cos \left (c+d\,x\right )+2\,d^2\,f^2\,x^2\,\cos \left (2\,c+2\,d\,x\right )-2\,d\,f^2\,x\,\sin \left (2\,c+2\,d\,x\right )+16\,d\,e\,f\,\cos \left (c+d\,x\right )+4\,d^2\,e\,f\,x\,\cos \left (2\,c+2\,d\,x\right )+16\,d^2\,e\,f\,x\,\sin \left (c+d\,x\right )}{8\,a\,d^3} \]
(8*d^2*e^2*sin(c + d*x) - f^2*cos(2*c + 2*d*x) - 16*f^2*sin(c + d*x) + 2*d ^2*e^2*cos(2*c + 2*d*x) + 8*d^2*f^2*x^2*sin(c + d*x) - 2*d*e*f*sin(2*c + 2 *d*x) + 16*d*f^2*x*cos(c + d*x) + 2*d^2*f^2*x^2*cos(2*c + 2*d*x) - 2*d*f^2 *x*sin(2*c + 2*d*x) + 16*d*e*f*cos(c + d*x) + 4*d^2*e*f*x*cos(2*c + 2*d*x) + 16*d^2*e*f*x*sin(c + d*x))/(8*a*d^3)