Integrand size = 25, antiderivative size = 196 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\frac {(3 a-4 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {(3 a-4 b) b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 a f}-\frac {(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f} \]
-1/3*(3*a-4*b)*cos(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2)/a/f+2/3*cos(f*x+e)^3*(a +b*sec(f*x+e)^2)^(5/2)/a/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(5/2)/a/f+1 /2*(3*a-4*b)*arctanh(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/ f+1/2*(3*a-4*b)*b*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/a/f
Time = 1.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\frac {\sqrt {2} \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-\frac {6 b \left (a+b-a \sin ^2(e+f x)\right )^{5/2}}{a}+15 \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )^{5/2}-5 (3 a-4 b) \left (-3 b^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )+\sqrt {a+b-a \sin ^2(e+f x)} \left (a+4 b-a \sin ^2(e+f x)\right )\right )\right )}{15 b f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \]
(Sqrt[2]*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*((-6*b*(a + b - a*Sin [e + f*x]^2)^(5/2))/a + 15*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)^(5/2) - 5*(3*a - 4*b)*(-3*b^(3/2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b ]] + Sqrt[a + b - a*Sin[e + f*x]^2]*(a + 4*b - a*Sin[e + f*x]^2))))/(15*b* f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))
Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4622, 365, 27, 359, 247, 211, 224, 219}
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.
| \(\displaystyle \int \sin ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4622 |
| \(\displaystyle \frac {\int \cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int -5 a \cos ^4(e+f x) \left (2-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\int \cos ^4(e+f x) \left (2-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{3 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \left (3 b \int \sqrt {b \sec ^2(e+f x)+a}d\sec (e+f x)-\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}\right )}{3 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \left (3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}\right )}{3 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \left (3 b \left (\frac {1}{2} a \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}\right )}{3 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \left (3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}\right )}{3 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a}}{f}\) |
((2*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(5/2))/(3*a) - (Cos[e + f*x]^5*( a + b*Sec[e + f*x]^2)^(5/2))/(5*a) + ((3*a - 4*b)*(-(Cos[e + f*x]*(a + b*S ec[e + f*x]^2)^(3/2)) + 3*b*((a*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b* Sec[e + f*x]^2]])/(2*Sqrt[b]) + (Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/ 2)))/(3*a))/f
3.1.80.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Leaf count of result is larger than twice the leaf count of optimal. \(801\) vs. \(2(172)=344\).
Time = 7.03 (sec) , antiderivative size = 802, normalized size of antiderivative = 4.09
| method | result | size |
| default | \(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (6 \cos \left (f x +e \right )^{8} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b +6 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \cos \left (f x +e \right )^{7}+60 \cos \left (f x +e \right )^{3} b^{\frac {5}{2}} \ln \left (-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )-4 \sec \left (f x +e \right ) b \right ) a -20 \cos \left (f x +e \right )^{6} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b +12 \cos \left (f x +e \right )^{6} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2}-45 \cos \left (f x +e \right )^{3} b^{\frac {3}{2}} \ln \left (-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )-4 \sec \left (f x +e \right ) b \right ) a^{2}-20 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \cos \left (f x +e \right )^{5}+12 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2} \cos \left (f x +e \right )^{5}+30 \cos \left (f x +e \right )^{4} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b -80 \cos \left (f x +e \right )^{4} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2}+6 \cos \left (f x +e \right )^{4} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3}+30 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )^{3} a^{2} b -80 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )^{3} b^{2} a +6 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \cos \left (f x +e \right )^{3}-15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )^{2} b^{2} a -15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) a \,b^{2}\right )}{30 f a b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (1+\cos \left (f x +e \right )\right )}\) | \(802\) |
-1/30/f/a/b*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) ^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))*(6*cos(f*x+e)^8*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b+6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 1/2)*a^2*b*cos(f*x+e)^7+60*cos(f*x+e)^3*b^(5/2)*ln(-4*b^(1/2)*((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)*sec(f*x+e)-4*sec(f*x+e)*b)*a-20*cos(f*x+e)^6*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b+12*cos(f*x+e)^6*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*a*b^2-45*cos(f*x+e)^3*b^(3/2)*ln(-4*b^(1/2)*((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*sec(f*x+e)-4*sec(f*x+e)*b)*a^2-20*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*a^2*b*cos(f*x+e)^5+12*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*a*b^2*cos(f*x+e)^5+30*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x +e))^2)^(1/2)*a^2*b-80*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*a*b^2+6*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3 +30*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^3*a^2*b-80*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^3*b^2*a+6*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*cos(f*x+e)^3-15*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*b^2*a-15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*cos(f*x+e)*a*b^2)
Time = 0.53 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.79 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\left [-\frac {15 \, {\left (3 \, a^{2} - 4 \, a b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (6 \, a^{2} \cos \left (f x + e\right )^{6} - 4 \, {\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (15 \, a^{2} - 40 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, a f \cos \left (f x + e\right )}, -\frac {15 \, {\left (3 \, a^{2} - 4 \, a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) + {\left (6 \, a^{2} \cos \left (f x + e\right )^{6} - 4 \, {\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (15 \, a^{2} - 40 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, a f \cos \left (f x + e\right )}\right ] \]
[-1/60*(15*(3*a^2 - 4*a*b)*sqrt(b)*cos(f*x + e)*log((a*cos(f*x + e)^2 - 2* sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/co s(f*x + e)^2) + 2*(6*a^2*cos(f*x + e)^6 - 4*(5*a^2 - 3*a*b)*cos(f*x + e)^4 + 2*(15*a^2 - 40*a*b + 3*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)), -1/30*(15*(3*a^2 - 4*a*b)*s qrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f* x + e)/b)*cos(f*x + e) + (6*a^2*cos(f*x + e)^6 - 4*(5*a^2 - 3*a*b)*cos(f*x + e)^4 + 2*(15*a^2 - 40*a*b + 3*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt((a*cos (f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*cos(f*x + e))]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.41 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=-\frac {\frac {12 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5}}{a} - 40 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} + 60 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a \cos \left (f x + e\right ) - 120 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) - \frac {30 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a b \cos \left (f x + e\right )}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + 45 \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - 60 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{60 \, f} \]
-1/60*(12*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5/a - 40*(a + b/cos(f* x + e)^2)^(3/2)*cos(f*x + e)^3 + 60*sqrt(a + b/cos(f*x + e)^2)*a*cos(f*x + e) - 120*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e) - 30*sqrt(a + b/cos(f* x + e)^2)*a*b*cos(f*x + e)/((a + b/cos(f*x + e)^2)*cos(f*x + e)^2 - b) + 4 5*a*sqrt(b)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt( a + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))) - 60*b^(3/2)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a + b/cos(f*x + e)^2)*cos( f*x + e) + sqrt(b))))/f
Leaf count of result is larger than twice the leaf count of optimal. 1952 vs. \(2 (172) = 344\).
Time = 1.95 (sec) , antiderivative size = 1952, normalized size of antiderivative = 9.96 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\text {Too large to display} \]
-1/15*(15*(3*a*b - 4*b^2)*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f *x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b))/sqrt(-b ))*sgn(cos(f*x + e))/sqrt(-b) + 30*((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^3*(a*b + 2*b^2)*sgn(cos (f*x + e)) - (sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/ 2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1 /2*f*x + 1/2*e)^2 + a + b))^2*(3*a*b - 2*b^2)*sqrt(a + b)*sgn(cos(f*x + e) ) + (3*a^2*b - 3*a*b^2 - 2*b^3)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt (a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1 /2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sgn(cos(f*x + e)) - (a^2*b - a*b^2 + 2*b^3)*sqrt(a + b)*sgn(cos(f*x + e)))/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2* a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - 2*(sqr t(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/ 2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) + a - 3*b)^2 - 4*(15*(sqrt(a + b)*tan(1/2*f*x + 1/2 *e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan (1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^9*(2*a*b - b...
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]