Integrand size = 23, antiderivative size = 122 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 f}-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac {b \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f} \]
-a^(3/2)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f-1/2*(3*a +b)*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))*b^(1/2)/f-1/2*b* cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f
Time = 0.68 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.16 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {4 a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\sqrt {2} b \cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))}-2 \sqrt {-b} (3 a+b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{4 f} \]
-1/4*(4*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Co s[2*(e + f*x)]]] + Sqrt[2]*b*Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x) ]] - 2*Sqrt[-b]*(3*a + b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/f
Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3665, 318, 25, 398, 224, 216, 291, 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.
| \(\displaystyle \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \frac {\left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle -\frac {\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}-\frac {1}{2} \int -\frac {(a+b) (2 a+b)-b (3 a+b) \cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {1}{2} \int \frac {(a+b) (2 a+b)-b (3 a+b) \cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+b (3 a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)\right )+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+b (3 a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}\right )+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )\right )+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )\right )+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {1}{2} \left (2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )\right )+\frac {1}{2} b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
-(((Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]] + 2*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/2 + (b*Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/2)/f)
3.2.34.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[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. \(254\) vs. \(2(104)=208\).
Time = 1.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.09
| method | result | size |
| default | \(\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (b^{\frac {3}{2}} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )-2 a^{\frac {3}{2}} \ln \left (\frac {-\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}-a -b}{\cos ^{2}\left (f x +e \right )-1}\right )+3 \sqrt {b}\, \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a -2 b \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\right )}{4 \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(255\) |
1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(b^(3/2)*arctan(1/2*(-2*b*cos( f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))-2*a^(3/2 )*ln((-(a-b)*cos(f*x+e)^2-2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^( 1/2)-a-b)/(cos(f*x+e)^2-1))+3*b^(1/2)*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b ^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))*a-2*b*(-b*cos(f*x+e)^4+ (a+b)*cos(f*x+e)^2)^(1/2))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (104) = 208\).
Time = 0.58 (sec) , antiderivative size = 1282, normalized size of antiderivative = 10.51 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[-1/16*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) - (3*a + b)*sqrt( -b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a ^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b ^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b ^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt (-b*cos(f*x + e)^2 + a + b)*sqrt(-b)) - 4*a^(3/2)*log(2*((a^2 - 6*a*b + b^ 2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 - 4*((a - b)*co s(f*x + e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt( a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/f, 1/16* (8*sqrt(-a)*a*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e))) - 8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) + (3*a + b)*sqrt(-b)*l og(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^ 2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b^3*co s(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3 )*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*c os(f*x + e)^2 + a + b)*sqrt(-b)))/f, 1/8*((3*a + b)*sqrt(b)*arctan(1/4*(8* b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*...
\[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \csc {\left (e + f x \right )}\, dx \]
Time = 0.50 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.47 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 \, a \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right ) + b^{\frac {3}{2}} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) + a^{\frac {3}{2}} \log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right ) - a^{\frac {3}{2}} \log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, f} \]
-1/2*(3*a*sqrt(b)*arcsin(b*cos(f*x + e)/sqrt(a*b + b^2)) + b^(3/2)*arcsin( b*cos(f*x + e)/sqrt(a*b + b^2)) + sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f* x + e) + a^(3/2)*log(b - sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a)/(cos(f*x + e) - 1) - a/(cos(f*x + e) - 1)) - a^(3/2)*log(-b + sqrt(-b*cos(f*x + e)^ 2 + a + b)*sqrt(a)/(cos(f*x + e) + 1) + a/(cos(f*x + e) + 1)))/f
Exception generated. \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{\sin \left (e+f\,x\right )} \,d x \]