Integrand size = 25, antiderivative size = 104 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}-\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}-\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b f} \]
a^(3/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f-1/3*(a+b*sec(f*x+e)^2) ^(3/2)/f+1/5*(a+b*sec(f*x+e)^2)^(5/2)/b/f-a*(a+b*sec(f*x+e)^2)^(1/2)/f
Time = 1.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {15 a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )-15 a b \sqrt {a+b \sec ^2(e+f x)}-5 b \left (a+b \sec ^2(e+f x)\right )^{3/2}+3 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{15 b f} \]
(15*a^(3/2)*b*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] - 15*a*b*Sqrt[a + b*Sec[e + f*x]^2] - 5*b*(a + b*Sec[e + f*x]^2)^(3/2) + 3*(a + b*Sec[e + f*x]^2)^(5/2))/(15*b*f)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4627, 25, 354, 90, 60, 60, 73, 221}
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.
\(\displaystyle \int \tan ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \frac {\int -\cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec ^2(e+f x)-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {a \int \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec ^2(e+f x)-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {a \left (a \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+2 \sqrt {a+b \sec ^2(e+f x)}\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a \left (\frac {2 a \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sec ^2(e+f x)}\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a \left (2 \sqrt {a+b \sec ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\) |
-1/2*((2*(a + b*Sec[e + f*x]^2)^(3/2))/3 - (2*(a + b*Sec[e + f*x]^2)^(5/2) )/(5*b) + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*Sq rt[a + b*Sec[e + f*x]^2]))/f
3.4.90.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), 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] || IGtQ[p, 0] || Integers Q[2*n, p])
Leaf count of result is larger than twice the leaf count of optimal. \(517\) vs. \(2(88)=176\).
Time = 14.22 (sec) , antiderivative size = 518, normalized size of antiderivative = 4.98
method | result | size |
default | \(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (15 \cos \left (f x +e \right )^{3} a^{\frac {3}{2}} \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 \cos \left (f x +e \right ) a +4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) b +3 \cos \left (f x +e \right )^{3} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}-20 \cos \left (f x +e \right )^{3} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +3 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}-20 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +6 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -5 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \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 -5 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+3 b^{2} \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 )+3 b^{2} \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 )^{2}\right )}{15 f 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 )}\) | \(518\) |
1/15/f/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))*(15*cos(f*x+e)^3*a^(3/2)*ln(4*cos(f* x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+4* a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*b+3*cos(f*x+e)^3*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-20*cos(f*x+e)^3*((b+a*cos(f*x+ e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+3*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*a^2-20*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*a*b+6*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b- 5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)+6*((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*b^2+3*b^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)+ 3*b^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)^2)
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (88) = 176\).
Time = 1.97 (sec) , antiderivative size = 443, normalized size of antiderivative = 4.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b \cos \left (f x + e\right )^{4} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, b f \cos \left (f x + e\right )^{4}}, -\frac {15 \, \sqrt {-a} a b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{4} - 4 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, b f \cos \left (f x + e\right )^{4}}\right ] \]
[1/120*(15*a^(3/2)*b*cos(f*x + e)^4*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b *cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b ^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + 8*((3*a^2 - 20*a*b)*cos(f*x + e)^4 + (6*a*b - 5*b^2)*cos(f*x + e)^2 + 3*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^4), -1/60*(15*sqrt(-a)*a*b*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos (f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2 *a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^4 - 4* ((3*a^2 - 20*a*b)*cos(f*x + e)^4 + (6*a*b - 5*b^2)*cos(f*x + e)^2 + 3*b^2) *sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^4)]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1431 vs. \(2 (88) = 176\).
Time = 1.67 (sec) , antiderivative size = 1431, normalized size of antiderivative = 13.76 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Too large to display} \]
-2/15*(15*a^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(-a))*sgn(cos(f *x + e))/sqrt(-a) - 2*(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*a^2*sgn(cos(f*x + e)) - 15*(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))^8*(7*a^2 - 8*a*b - 4*b^2)*sqrt(a + b)*sgn(cos(f*x + e)) + 20*(1 5*a^3 - 21*a^2*b - 12*a*b^2 + 8*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))^7*sgn(cos(f*x + e)) - 20*(21*a^3 - 63*a^2*b + 60*a*b^2 - 4*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))^6*sqrt(a + b)*sgn (cos(f*x + e)) + 2*(105*a^4 - 630*a^3*b + 1065*a^2*b^2 + 360*a*b^3 - 16*b^ 4)*(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))^5*sgn(cos(f*x + e)) + 10*(21*a^4 + 42*a^3*b - 303*a^2*b^ 2 + 336*a*b^3 + 4*b^4)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan...
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]