Integrand size = 25, antiderivative size = 193 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {(5 a-b) (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{3/2} f}+\frac {(5 a-b) (a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac {\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f} \]
1/16*(5*a-b)*(a+b)^2*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2)) /a^(3/2)/f+1/16*(5*a-b)*(a+b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^( 1/2)/a/f+1/24*(5*a-b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/a /f+1/6*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(5/2)/a/f
Time = 2.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.85 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\frac {3 \sqrt {2} \sqrt {a+b} \left (5 a^2+4 a b-b^2\right ) \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}}}+\sqrt {a} \left (23 a^2+29 a b+3 b^2+a (9 a+7 b) \cos (2 (e+f x))+a^2 \cos (4 (e+f x))\right ) \sin (e+f x)\right )}{48 a^{3/2} f} \]
(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*((3*Sqrt[2]*Sqrt[a + b]*(5*a^2 + 4*a*b - b^2)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/Sqrt[(a + 2*b + a *Cos[2*(e + f*x)])/(a + b)] + Sqrt[a]*(23*a^2 + 29*a*b + 3*b^2 + a*(9*a + 7*b)*Cos[2*(e + f*x)] + a^2*Cos[4*(e + f*x)])*Sin[e + f*x]))/(48*a^(3/2)*f )
Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4634, 296, 292, 292, 291, 216}
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.
| \(\displaystyle \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sec (e+f x)^6}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 296 |
| \(\displaystyle \frac {\frac {(5 a-b) \int \frac {\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)}{6 a}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {\frac {(5 a-b) \left (\frac {3}{4} (a+b) \int \frac {\sqrt {b \tan ^2(e+f x)+a+b}}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )}{6 a}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {\frac {(5 a-b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )}{6 a}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {(5 a-b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )}{6 a}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {(5 a-b) \left (\frac {3}{4} (a+b) \left (\frac {(a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {a}}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )}{6 a}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*a*(1 + Tan[e + f*x]^2) ^3) + ((5*a - b)*((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*(1 + Tan[e + f*x]^2)^2) + (3*(a + b)*(((a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sq rt[a + b + b*Tan[e + f*x]^2]])/(2*Sqrt[a]) + (Tan[e + f*x]*Sqrt[a + b + b* Tan[e + f*x]^2])/(2*(1 + Tan[e + f*x]^2))))/4))/(6*a))/f
3.3.53.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(985\) vs. \(2(173)=346\).
Time = 6.04 (sec) , antiderivative size = 986, normalized size of antiderivative = 5.11
1/48/f/a/(-a)^(1/2)*(8*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 /2)*a^2*cos(f*x+e)^5*sin(f*x+e)+8*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^4*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+10*(-a)^(1/2)*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)^3*sin(f*x+e)+14*(-a)^(1/2)*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*cos(f*x+e)^3*sin(f*x+e)+10*(-a) ^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) *a^2+14*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*a*b+15*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) *a^2*cos(f*x+e)*sin(f*x+e)+22*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) )^2)^(1/2)*a*b*cos(f*x+e)*sin(f*x+e)+3*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c os(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)*sin(f*x+e)+15*(-a)^(1/2)*sin(f*x+e)*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+22*(-a)^(1/2)*sin(f*x+e)*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+3*(-a)^(1/2)*sin(f*x+e)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+15*ln(4*(-a)^(1/2)*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^3+27*ln(4*(-a)^(1/2)*((b+a*cos( f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e) ^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^2*b+9*ln(4*(-a)^(1/2)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^2-3*ln(4*(-a)^(1/2)...
Time = 1.19 (sec) , antiderivative size = 647, normalized size of antiderivative = 3.35 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (5 \, a^{3} + 9 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} + 22 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{384 \, a^{2} f}, -\frac {3 \, {\left (5 \, a^{3} + 9 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} + 22 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, a^{2} f}\right ] \]
[1/384*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^ 2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*c os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*a^3*cos(f*x + e)^5 + 2*(5*a^3 + 7*a^2*b)*cos(f*x + e)^3 + (15*a^3 + 22*a^2*b + 3*a*b^2)*cos( f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f ), -1/192*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*sqrt(a)*arctan(1/4*(8*a^2*c os(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(8*a^3*cos(f*x + e)^5 + 2*(5*a^3 + 7*a^2*b)*cos(f*x + e)^3 + (15*a^3 + 22*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) ^2)*sin(f*x + e))/(a^2*f)]
Timed out. \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{6} \,d x } \]
\[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]