Integrand size = 23, antiderivative size = 88 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\frac {11 a^2 \cos (e+f x)}{3 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec (e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f} \] Output:
11/3*a^2*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)+7/3*sec(f*x+e)*(a+a*sin(f*x+e ))^(3/2)/f-2/3*sec(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f
Time = 5.71 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\frac {a \sec (e+f x) (15+\cos (2 (e+f x))-8 \sin (e+f x)) \sqrt {a (1+\sin (e+f x))}}{3 f} \] Input:
Integrate[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^2,x]
Output:
(a*Sec[e + f*x]*(15 + Cos[2*(e + f*x)] - 8*Sin[e + f*x])*Sqrt[a*(1 + Sin[e + f*x])])/(3*f)
Time = 0.51 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3192, 27, 3042, 3334, 3042, 3125}
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.
\(\displaystyle \int \tan ^2(e+f x) (a \sin (e+f x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^2 (a \sin (e+f x)+a)^{3/2}dx\) |
\(\Big \downarrow \) 3192 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec ^2(e+f x) (\sin (e+f x) a+a)^{3/2} (2 \sin (e+f x) a+5 a)dx}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec ^2(e+f x) (\sin (e+f x) a+a)^{3/2} (2 \sin (e+f x) a+5 a)dx}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{3/2} (2 \sin (e+f x) a+5 a)}{\cos (e+f x)^2}dx}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
\(\Big \downarrow \) 3334 |
\(\displaystyle \frac {\frac {7 a \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{f}-\frac {11}{2} a^2 \int \sqrt {\sin (e+f x) a+a}dx}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7 a \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{f}-\frac {11}{2} a^2 \int \sqrt {\sin (e+f x) a+a}dx}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {11 a^3 \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}+\frac {7 a \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{f}}{3 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^2,x]
Output:
(-2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*a*f) + ((11*a^3*Cos[e + f* x])/(f*Sqrt[a + a*Sin[e + f*x]]) + (7*a*Sec[e + f*x]*(a + a*Sin[e + f*x])^ (3/2))/f)/(3*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + Simp[1/(b*m) Int[(a + b*Sin[e + f*x])^m*((b*(m + 1) + a*Sin[e + f*x])/Cos [e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !LtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62
method | result | size |
default | \(-\frac {2 a^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}+4 \sin \left (f x +e \right )-8\right )}{3 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(55\) |
Input:
int((a+sin(f*x+e)*a)^(3/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
-2/3*a^2*(1+sin(f*x+e))*(sin(f*x+e)^2+4*sin(f*x+e)-8)/cos(f*x+e)/(a+sin(f* x+e)*a)^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.55 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 4 \, a \sin \left (f x + e\right ) + 7 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{3 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^2,x, algorithm="fricas")
Output:
2/3*(a*cos(f*x + e)^2 - 4*a*sin(f*x + e) + 7*a)*sqrt(a*sin(f*x + e) + a)/( f*cos(f*x + e))
\[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate((a+a*sin(f*x+e))**(3/2)*tan(f*x+e)**2,x)
Output:
Integral((a*(sin(e + f*x) + 1))**(3/2)*tan(e + f*x)**2, x)
Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.65 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=-\frac {8 \, {\left (2 \, a^{\frac {3}{2}} - \frac {2 \, a^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {2 \, a^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{3 \, f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^2,x, algorithm="maxima")
Output:
-8/3*(2*a^(3/2) - 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^(3/2)*si n(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 2*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)/(f*(sin(f*x + e)/ (cos(f*x + e) + 1) - 1)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (76) = 152\).
Time = 0.39 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.38 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=-\frac {\sqrt {2} {\left (3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{8} + 60 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{6} + 50 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + 60 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + 3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{6 \, {\left (\tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{7} + 3 \, \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{5} + 3 \, \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{3} + \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )\right )} f} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^2,x, algorithm="giac")
Output:
-1/6*sqrt(2)*(3*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f* x + 1/4*e)^8 + 60*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4* f*x + 1/4*e)^6 + 50*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/ 4*f*x + 1/4*e)^4 + 60*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 3*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/((ta n(-1/8*pi + 1/4*f*x + 1/4*e)^7 + 3*tan(-1/8*pi + 1/4*f*x + 1/4*e)^5 + 3*ta n(-1/8*pi + 1/4*f*x + 1/4*e)^3 + tan(-1/8*pi + 1/4*f*x + 1/4*e))*f)
Timed out. \[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:
int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(3/2),x)
Output:
int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(3/2), x)
\[ \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right ) \tan \left (f x +e \right )^{2}d x \right ) \] Input:
int((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^2,x)
Output:
sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1)*tan(e + f*x)**2,x) + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)*tan(e + f*x)**2,x))