Integrand size = 23, antiderivative size = 84 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f}-\frac {(a+b) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f} \]
(a+b)^(3/2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/f-1/3*(a+b*sin(f *x+e)^2)^(3/2)/f-(a+b)*(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {3 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b-b \cos ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b-b \cos ^2(e+f x)} \left (-4 (a+b)+b \cos ^2(e+f x)\right )}{3 f} \]
(3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b - b*Cos[e + f*x]^2]/Sqrt[a + b]] + Sqr t[a + b - b*Cos[e + f*x]^2]*(-4*(a + b) + b*Cos[e + f*x]^2))/(3*f)
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3673, 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.
\(\displaystyle \int \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x) \left (a+b \sin (e+f x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b) \left ((a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b) \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
((-2*(a + b*Sin[e + f*x]^2)^(3/2))/3 + (a + b)*(2*Sqrt[a + b]*ArcTanh[Sqrt [a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*Sin[e + f*x]^2]))/(2*f)
3.6.1.3.1 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_)^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. \(210\) vs. \(2(72)=144\).
Time = 1.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.51
method | result | size |
default | \(\frac {\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (b \left (\cos ^{2}\left (f x +e \right )\right )+2 a -b \right )}{3}-b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-2 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}+\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\sqrt {a +b}}+\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\sqrt {a +b}}}{f}\) | \(211\) |
(1/3*(a+b-b*cos(f*x+e)^2)^(1/2)*(b*cos(f*x+e)^2+2*a-b)-b*(a+b*sin(f*x+e)^2 )^(1/2)-2*a*(a+b*sin(f*x+e)^2)^(1/2)+(1/2*a^2+a*b+1/2*b^2)/(a+b)^(1/2)*ln( (2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e )))+(1/2*a^2+a*b+1/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^ 2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1)))/f
Time = 0.49 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.21 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\left [\frac {3 \, {\left (a + b\right )}^{\frac {3}{2}} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (b \cos \left (f x + e\right )^{2} - 4 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f}, -\frac {3 \, {\left (a + b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - {\left (b \cos \left (f x + e\right )^{2} - 4 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, f}\right ] \]
[1/6*(3*(a + b)^(3/2)*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*(b*cos(f*x + e)^2 - 4*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f, -1/3*(3*(a + b)*sqrt(-a - b)*a rctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - (b*cos(f*x + e)^2 - 4*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f]
\[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (72) = 144\).
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.87 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=-\frac {3 \, {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - 3 \, {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right ) + 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} + 6 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a + 6 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{6 \, f} \]
-1/6*(3*(a + b)^(3/2)*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1) ) - a/(sqrt(a*b)*(sin(f*x + e) + 1))) - 3*(a + b)^(3/2)*arcsinh(-b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1))) + 2*(b*sin(f*x + e)^2 + a)^(3/2) + 6*sqrt(b*sin(f*x + e)^2 + a)*a + 6*sqrt( b*sin(f*x + e)^2 + a)*b)/f
Leaf count of result is larger than twice the leaf count of optimal. 1270 vs. \(2 (72) = 144\).
Time = 0.71 (sec) , antiderivative size = 1270, normalized size of antiderivative = 15.12 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\text {Too large to display} \]
-2/3*(3*(a^2 + 2*a*b + b^2)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f *x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/sqrt(-a - b) - 2*(6*(sqrt(a)*t an(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a*b + 3*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*b^2 + 18*(sqrt(a)*tan(1/2*f*x + 1/2*e) ^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan( 1/2*f*x + 1/2*e)^2 + a))^4*a^(3/2)*b + 21*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2 *f*x + 1/2*e)^2 + a))^4*sqrt(a)*b^2 + 12*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2* f*x + 1/2*e)^2 + a))^3*a^2*b + 54*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a *tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1 /2*e)^2 + a))^3*a*b^2 + 40*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/ 2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^3 - 12*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1 /2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2* a^(5/2)*b + 18*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2* e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*...
Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int \mathrm {tan}\left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]