Integrand size = 25, antiderivative size = 113 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f} \]
1/3*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+arctanh(sec(f*x+e)*b^( 1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-cos(f*x+e)*(a-b+b*sec(f*x+e)^2) ^(1/2)/f
Time = 1.47 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\cos (e+f x) \left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a+b+(a-b) \cos (2 (e+f x))} (-5 a+7 b+(a-b) \cos (2 (e+f x)))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b) f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]
(Cos[e + f*x]*(6*Sqrt[2]*(a - b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[ 2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]* (-5*a + 7*b + (a - b)*Cos[2*(e + f*x)]))*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(6*Sqrt[2]*(a - b)*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4147, 25, 358, 247, 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.
| \(\displaystyle \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^3 \sqrt {a+b \tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {\int \cos ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {b \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\) |
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]] - Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2] + (Cos[e + f*x]^3*(a - b + b*S ec[e + f*x]^2)^(3/2))/(3*(a - b)))/f
3.1.93.3.1 Defintions of rubi rules used
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] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, -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. \(642\) vs. \(2(101)=202\).
Time = 0.50 (sec) , antiderivative size = 643, normalized size of antiderivative = 5.69
| method | result | size |
| default | \(\frac {\left (\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{3} a -\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cos \left (f x +e \right )^{3}+\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{2} a -\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{2} b -3 \ln \left (-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) b^{\frac {3}{2}}-3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right ) a +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cos \left (f x +e \right )+3 \ln \left (-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) \sqrt {b}\, a -3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )}{3 f \left (a -b \right ) \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(643\) |
1/3/f/(a-b)*(((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*co s(f*x+e)^3*a-((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b* 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)*co s(f*x+e)^2*a-((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*co s(f*x+e)^2*b-3*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)-3*((a*cos(f*x+e)^2-b*cos(f*x+e )^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)*a+4*((a*cos(f*x+e)^2-b*cos(f*x+e )^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b*cos(f*x+e)+3*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^(1/2) *a-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a+4*((a*co s(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b)*(a+b*tan(f*x+e)^2) ^(1/2)*cos(f*x+e)/(cos(f*x+e)+1)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f *x+e)+1)^2)^(1/2)
Time = 0.40 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.45 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a - b\right )} \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left (a - b\right )} f}, -\frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a - b\right )} f}\right ] \]
[1/6*(3*(a - b)*sqrt(b)*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*((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a - b)*c os(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f), -1/3*(3*(a - 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) - ((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a - b )*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f)]
Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\frac {2 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 3 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{6 \, f} \]
1/6*(2*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/(a - b) - 6*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - 3*sqrt(b)*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))))/f
Leaf count of result is larger than twice the leaf count of optimal. 1182 vs. \(2 (101) = 202\).
Time = 0.79 (sec) , antiderivative size = 1182, normalized size of antiderivative = 10.46 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]
2/3*(3*b*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) - 2*(3*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqr t(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 - 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))^4*a^(3/2) + 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 + 16*(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 - 50*(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 + 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^2 + 24*(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) - 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))^2*a^(3/2)*b + 24*(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 - 48*(sqrt(a)*tan(1/...
Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int {\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]