Integrand size = 23, antiderivative size = 113 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {3 (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f} \]
-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/f+3/2*(a-b)*arctanh(sec(f*x+e)*b^(1 /2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+3/2*b*sec(f*x+e)*(a-b+b*sec(f*x+ e)^2)^(1/2)/f
Time = 1.72 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right ) \cos ^2(e+f x)-2 (a-2 b+(a-b) \cos (2 (e+f x))) \sqrt {a+b+(a-b) \cos (2 (e+f x))}\right ) \sec (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{4 \sqrt {2} f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]
((6*Sqrt[2]*(a - b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]] /(Sqrt[2]*Sqrt[b])]*Cos[e + f*x]^2 - 2*(a - 2*b + (a - b)*Cos[2*(e + f*x)] )*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])*Sec[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(4*Sqrt[2]*f*Sqrt[a + b + (a - b)*Co s[2*(e + f*x)]])
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4147, 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.
\(\displaystyle \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x) \left (a+b \tan (e+f x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \frac {\int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {3 b \int \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {3 b \left (\frac {1}{2} (a-b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {3 b \left (\frac {1}{2} (a-b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 b \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}\) |
(-(Cos[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2)) + 3*b*(((a - b)*ArcTanh[ (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))/f
3.2.6.3.1 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[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. \(919\) vs. \(2(99)=198\).
Time = 2.49 (sec) , antiderivative size = 920, normalized size of antiderivative = 8.14
-1/2/f/b/(a-b)*cos(f*x+e)*(-3*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^(7/2)+6* 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^(5/2)*a-3*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*co s(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^2+2*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-4*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^2+2*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)*b^3+2*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^2*b-4*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*b^2+2*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)*b^3-cos(f*x+e)*((a*cos(f*x+e )^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a*b^2+cos(f*x+e)*((a*cos(f*x +e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b^3-a*((a*cos(f*x+e)^2-b*c os(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b^2+((a*cos(f*x+e)^2-b*cos(f*x+e)^2 +b)/(cos(f*x+e)+1)^2)^(1/2)*b^3)*(a+b*tan(f*x+e)^2)^(3/2)/((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)+1)/(a*cos(f*x+e)...
Time = 0.48 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.37 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a - b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) + {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \]
[-1/4*(3*(a - b)*sqrt(b)*cos(f*x + e)*log(-((a - b)*cos(f*x + e)^2 - 2*sqr t(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)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/2*(3*(a - b)*sqrt(-b)*ar ctan(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)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]
\[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sin {\left (e + f x \right )}\, dx \]
Time = 0.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {4 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) - \frac {2 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + \frac {3 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{4 \, f} \]
-1/4*(4*sqrt(a - b + b/cos(f*x + e)^2)*(a - b)*cos(f*x + e) - 2*(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) + 3*(a*b - b^2)*log((sqrt(a - b + b/cos(f*x + e)^2)*co s(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. 1366 vs. \(2 (99) = 198\).
Time = 1.88 (sec) , antiderivative size = 1366, normalized size of antiderivative = 12.09 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
(3*(a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 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) - sq rt(a))/sqrt(-b))/sqrt(-b) + 4*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta n(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*sgn(tan(1/2*f*x + 1/2*e)^2 - 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))*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))*b^2*sgn(tan(1/2*f*x + 1/2 *e)^2 - 1) - a^(5/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 2*a^(3/2)*b*sgn(tan (1/2*f*x + 1/2*e)^2 - 1) - sqrt(a)*b^2*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))^2 + 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) - 3*a + 4*b) + 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))^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*...
Timed out. \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int \sin \left (e+f\,x\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]