Integrand size = 23, antiderivative size = 121 \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac {(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f} \]
(a+b)^(3/2)*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f-1/2 *(3*a+2*b)*arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))*b^(1/2)/f- 1/2*b*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.72 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.93 \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {2} b (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\sqrt {2} \left (4 a^2+5 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {2 a+2 b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )-2 \sqrt {b} \sqrt {a+b} \left (\sqrt {2} (3 a+2 b) \log \left (\sqrt {2 a+b-b \cos (2 (e+f x))}+\sqrt {2} \sqrt {b} \sin (e+f x)\right )+\sqrt {b} \sqrt {2 a+b-b \cos (2 (e+f x))} \sin (e+f x)\right )}{4 \sqrt {2} \sqrt {a+b} f} \]
(Sqrt[2]*b*(3*a + 2*b)*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + Sqrt[2]*(4*a^2 + 5*a*b + 2*b^2)*ArcTanh[(Sqr t[2*a + 2*b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] - 2*Sqrt[b] *Sqrt[a + b]*(Sqrt[2]*(3*a + 2*b)*Log[Sqrt[2*a + b - b*Cos[2*(e + f*x)]] + Sqrt[2]*Sqrt[b]*Sin[e + f*x]] + Sqrt[b]*Sqrt[2*a + b - b*Cos[2*(e + f*x)] ]*Sin[e + f*x]))/(4*Sqrt[2]*Sqrt[a + b]*f)
Time = 0.34 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3669, 318, 25, 398, 224, 219, 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.
| \(\displaystyle \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\cos (e+f x)}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{1-\sin ^2(e+f x)}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {-\frac {1}{2} \int -\frac {b (3 a+2 b) \sin ^2(e+f x)+a (2 a+b)}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {b (3 a+2 b) \sin ^2(e+f x)+a (2 a+b)}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (a+b)^2 \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)-b (3 a+2 b) \int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (a+b)^2 \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)-b (3 a+2 b) \int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}\right )-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (a+b)^2 \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)-\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (a+b)^2 \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}-\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )-\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )-\frac {1}{2} b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}\) |
((-(Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]) + 2*(a + b)^(3/2)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b* Sin[e + f*x]^2]])/2 - (b*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/2)/f
3.4.36.3.1 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[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[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[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. \(268\) vs. \(2(103)=206\).
Time = 1.31 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.22
| method | result | size |
| default | \(\frac {-b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )-b^{2} \left (\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 b^{\frac {3}{2}}}\right )-2 a \sqrt {b}\, \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )+\frac {\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\sqrt {a +b}}+\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\sqrt {a +b}}}{f}\) | \(269\) |
(-b^(3/2)*ln(b^(1/2)*sin(f*x+e)+(a+b*sin(f*x+e)^2)^(1/2))-b^2*(1/2*sin(f*x +e)/b*(a+b*sin(f*x+e)^2)^(1/2)-1/2*a/b^(3/2)*ln(b^(1/2)*sin(f*x+e)+(a+b*si n(f*x+e)^2)^(1/2)))-2*a*b^(1/2)*ln(b^(1/2)*sin(f*x+e)+(a+b*sin(f*x+e)^2)^( 1/2))+(-1/2*a^2-a*b-1/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+ e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))+(1/2*a^2+a*b+1/2*b^2)/(a+b )^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/( sin(f*x+e)-1)))/f
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (103) = 206\).
Time = 0.69 (sec) , antiderivative size = 1381, normalized size of antiderivative = 11.41 \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/16*((3*a + 2*b)*sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4 )*cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e)^2 + 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt( b)*sin(f*x + e)) + 4*(a + b)^(3/2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e) ^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8* a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) - 8*sqrt(-b*cos(f*x + e)^2 + a + b)* b*sin(f*x + e))/f, -1/16*(8*(a + b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos (f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a *b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e))) - (3*a + 2*b) *sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160* a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4 )*cos(f*x + e)^2 + 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^ 3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)*sin(f*x + e)) + 8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*sin(f*x + e))/f, 1/8*((3*a + 2*b)*...
\[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec {\left (e + f x \right )}\, dx \]
Time = 0.39 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.39 \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right ) + 2 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right ) - {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right ) + \sqrt {b \sin \left (f x + e\right )^{2} + a} b \sin \left (f x + e\right )}{2 \, f} \]
-1/2*(3*a*sqrt(b)*arcsinh(b*sin(f*x + e)/sqrt(a*b)) + 2*b^(3/2)*arcsinh(b* sin(f*x + e)/sqrt(a*b)) - (a + b)^(3/2)*arcsinh(b*sin(f*x + e)/(sqrt(a*b)* (sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1))) - (a + b)^(3/2)*ar csinh(-b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f *x + e) - 1))) + sqrt(b*sin(f*x + e)^2 + a)*b*sin(f*x + e))/f
Timed out. \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{\cos \left (e+f\,x\right )} \,d x \]