Integrand size = 25, antiderivative size = 118 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} f}+\frac {2 a-b}{2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]
-1/2*(2*a-b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f+1 /2*(2*a-b)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/2*sec(f*x+e)^2/(a+b)/f/(a+ b*sin(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(2 a-b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin ^2(e+f x)}{a+b}\right )+(a+b) \sec ^2(e+f x)}{2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
((2*a - b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[e + f*x]^2)/(a + b)] + (a + b)*Sec[e + f*x]^2)/(2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2])
Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3673, 87, 61, 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.
| \(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^3}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3673 |
| \(\displaystyle \frac {\int \frac {\sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right )^2 \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-b) \left (\frac {\int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a+b}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-b) \left (\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b (a+b)}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-b) \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{2 f}\) |
(1/((a + b)*(1 - Sin[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2]) - ((2*a - b)* ((2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b)^(3/2) - 2/((a + b)*Sqrt[a + b*Sin[e + f*x]^2])))/(2*(a + b)))/(2*f)
3.6.22.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(2536\) vs. \(2(102)=204\).
Time = 4.02 (sec) , antiderivative size = 2537, normalized size of antiderivative = 21.50
1/4/(cos(f*x+e)^4*a^3*b^2+3*cos(f*x+e)^4*a^2*b^3+3*cos(f*x+e)^4*a*b^4+cos( f*x+e)^4*b^5-2*cos(f*x+e)^2*a^4*b-8*cos(f*x+e)^2*a^3*b^2-12*cos(f*x+e)^2*a ^2*b^3-8*cos(f*x+e)^2*a*b^4-2*b^5*cos(f*x+e)^2+a^5+5*a^4*b+10*a^3*b^2+10*a ^2*b^3+5*a*b^4+b^5)/(a+b)^(1/2)/cos(f*x+e)^2*(2*(-b*cos(f*x+e)^2+(a*b^2+b^ 3)/b^2)^(3/2)*(a+b)^(1/2)*b^2*cos(f*x+e)^4-2*ln(2/(1+sin(f*x+e))*((a+b)^(1 /2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2*cos(f*x+e)^6-ln(2/ (1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a* b^3*cos(f*x+e)^6-2*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^( 1/2)+b*sin(f*x+e)+a))*a^2*b^2*cos(f*x+e)^6-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2 )*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^3*cos(f*x+e)^6-4*(a+b)^( 1/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b*cos(f*x+e)^2+2*(-b*cos(f*x+e)^2+(a*b^2 +b^3)/b^2)^(1/2)*(a+b)^(1/2)*a^2*b*cos(f*x+e)^2+10*(-b*cos(f*x+e)^2+(a*b^2 +b^3)/b^2)^(1/2)*(a+b)^(1/2)*a*b^2*cos(f*x+e)^2+6*(a+b)^(1/2)*(a+b-b*cos(f *x+e)^2)^(1/2)*a^2*b*cos(f*x+e)^2-2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2 )*(a+b)^(1/2)*a*b^2*cos(f*x+e)^6+4*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)* a*b^2*cos(f*x+e)^6+2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*(a+b)^(1/2)*a *b*cos(f*x+e)^4+4*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*a^2* b*cos(f*x+e)^4-8*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*a*b^2 *cos(f*x+e)^4+2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*b^2*cos(f*x+e)^4-12 *(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*b^3*cos(f*x+e)^4+4...
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (102) = 204\).
Time = 0.45 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.77 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left ({\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left ({\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - {\left ({\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left ({\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]
[-1/4*(((2*a*b - b^2)*cos(f*x + e)^4 - (2*a^2 + a*b - b^2)*cos(f*x + e)^2) *sqrt(a + b)*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqr t(a + b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*((2*a^2 + a*b - b^2)*cos(f*x + e )^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b + 3*a^2* b^2 + 3*a*b^3 + b^4)*f*cos(f*x + e)^4 - (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b ^3 + b^4)*f*cos(f*x + e)^2), 1/2*(((2*a*b - b^2)*cos(f*x + e)^4 - (2*a^2 + a*b - b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - ((2*a^2 + a*b - b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b + 3*a^2*b^2 + 3*a*b^ 3 + b^4)*f*cos(f*x + e)^4 - (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*f* cos(f*x + e)^2)]
\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.36 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.64 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {{\left (2 \, a b^{2} - b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {2 \, {\left (2 \, a^{2} b^{2} + 2 \, a b^{3} - {\left (2 \, a b^{2} - b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} {\left (a^{2} + 2 \, a b + b^{2}\right )} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}}{4 \, b^{2} f} \]
1/4*((2*a*b^2 - b^3)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt( b*sin(f*x + e)^2 + a) + sqrt(a + b)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) - 2*(2*a^2*b^2 + 2*a*b^3 - (2*a*b^2 - b^3)*(b*sin(f*x + e)^2 + a))/((b*sin(f *x + e)^2 + a)^(3/2)*(a^2 + 2*a*b + b^2) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3) *sqrt(b*sin(f*x + e)^2 + a)))/(b^2*f)
\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]