Integrand size = 25, antiderivative size = 186 \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {(3 a-5 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {(3 a-5 b) b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 (a-b) f}-\frac {(3 a-5 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f} \]
-1/3*(3*a-5*b)*cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+1/3*cos(f*x+e )^3*(a-b+b*sec(f*x+e)^2)^(5/2)/(a-b)/f+1/2*(3*a-5*b)*arctanh(sec(f*x+e)*b^ (1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*(3*a-5*b)*b*sec(f*x+e)*(a- b+b*sec(f*x+e)^2)^(1/2)/(a-b)/f
Time = 2.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.01 \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (12 \sqrt {2} (3 a-5 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right ) \cos ^2(e+f x)+\sqrt {a+b+(a-b) \cos (2 (e+f x))} (-9 a+37 b-8 (a-3 b) \cos (2 (e+f x))+(a-b) \cos (4 (e+f x)))\right ) \sec (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{24 \sqrt {2} f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]
((12*Sqrt[2]*(3*a - 5*b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f *x)]]/(Sqrt[2]*Sqrt[b])]*Cos[e + f*x]^2 + Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-9*a + 37*b - 8*(a - 3*b)*Cos[2*(e + f*x)] + (a - b)*Cos[4*(e + f* x)]))*Sec[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2] )/(24*Sqrt[2]*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])
Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 25, 359, 247, 211, 224, 219}
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.
| \(\displaystyle \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^3 \left (a+b \tan (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {(3 a-5 b) \int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{3 (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {(3 a-5 b) \left (3 b \int \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {(3 a-5 b) \left (3 b \left (\frac {1}{2} (a-b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {(3 a-5 b) \left (3 b \left (\frac {1}{2} (a-b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(3 a-5 b) \left (3 b \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\) |
((Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(5/2))/(3*(a - b)) + ((3*a - 5 *b)*(-(Cos[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2)) + 3*b*(((a - b)*ArcT anh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*Sqrt[b]) + (Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/2)))/(3*(a - b)))/f
3.2.5.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(1638\) vs. \(2(166)=332\).
Time = 2.68 (sec) , antiderivative size = 1639, normalized size of antiderivative = 8.81
1/6/f/b/(a-b)^2*(-15*cos(f*x+e)^2*ln(-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x +e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2 +b)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(9/2)-33*cos(f*x+ e)^2*ln(-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1 /2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*s ec(f*x+e)-4*b*sec(f*x+e))*b^(5/2)*a^2+9*cos(f*x+e)^2*ln(-4*b^(1/2)*((a*cos (f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+ e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))* b^(3/2)*a^3+2*cos(f*x+e)^5*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+ 1)^2)^(1/2)*a^3*b-6*cos(f*x+e)^5*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f *x+e)+1)^2)^(1/2)*a^2*b^2+6*cos(f*x+e)^5*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b )/(cos(f*x+e)+1)^2)^(1/2)*a*b^3+2*cos(f*x+e)^4*((a*cos(f*x+e)^2-b*cos(f*x+ e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^3*b-6*cos(f*x+e)^4*((a*cos(f*x+e)^2-b*co s(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^2+6*cos(f*x+e)^4*((a*cos(f*x+e )^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a*b^3-6*cos(f*x+e)^3*((a*cos (f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^3*b+26*cos(f*x+e)^3* ((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^2-34*cos( f*x+e)^3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a*b^3- 6*cos(f*x+e)^2*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)* a^3*b+26*cos(f*x+e)^2*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)...
Time = 0.49 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.65 \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (3 \, a - 5 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (3 \, a - 5 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) - {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \]
[-1/12*(3*(3*a - 5*b)*sqrt(b)*cos(f*x + e)*log(-((a - b)*cos(f*x + e)^2 - 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(2*(a - b)*cos(f*x + e)^4 - 2*(3*a - 7*b)*cos(f* x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos( f*x + e)), -1/6*(3*(3*a - 5*b)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos( f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) - (2*(a - b)* cos(f*x + e)^4 - 2*(3*a - 7*b)*cos(f*x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]
Timed out. \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.64 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.59 \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {4 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 12 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) + 12 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) + 6 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) + \frac {6 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} - \frac {9 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{12 \, f} \]
1/12*(4*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 12*sqrt(a - b + b/cos(f*x + e)^2)*(a - b)*cos(f*x + e) + 12*sqrt(a - b + b/cos(f*x + e)^2) *b*cos(f*x + e) + 6*b^(3/2)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))) + 6 *(a*b - b^2)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/((a - b + b/cos(f *x + e)^2)*cos(f*x + e)^2 - b) - 9*(a*b - b^2)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b)))/sqrt(b))/f
Leaf count of result is larger than twice the leaf count of optimal. 2688 vs. \(2 (166) = 332\).
Time = 2.96 (sec) , antiderivative size = 2688, normalized size of antiderivative = 14.45 \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/3*(3*(3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 5*b^2*sgn(tan(1/2*f*x + 1/ 2*e)^2 - 1))*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(-b))/sqrt(-b) + 6*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - s qrt(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*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + (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^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 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))^2*a^(3/2)*b*s gn(tan(1/2*f*x + 1/2*e)^2 - 1) + 5*(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*sqrt(a)*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 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))*a^2*b*sgn(tan(1/2*f*x + 1/2*e )^2 - 1) - 9*(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))*a*b^2*s gn(tan(1/2*f*x + 1/2*e)^2 - 1) + 4*(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))*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^(5/2)*b*sgn(tan(...
Timed out. \[ \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]