Integrand size = 25, antiderivative size = 127 \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}+\frac {(a-2 b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac {(a+b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{2 f} \]
b^(3/2)*arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f+1/2*(a-2*b) *arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))*(a+b)^(1/2)/f+1/ 2*(a+b)*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Time = 0.96 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.65 \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-2 b^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+2 \left (a^2-a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2 a+2 b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\sqrt {a+b} \left (4 b^{3/2} \log \left (\sqrt {2 a+b-b \cos (2 (e+f x))}+\sqrt {2} \sqrt {b} \sin (e+f x)\right )+\sqrt {2} (a+b) \sqrt {2 a+b-b \cos (2 (e+f x))} \sec (e+f x) \tan (e+f x)\right )}{4 \sqrt {a+b} f} \]
(-2*b^2*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2* (e + f*x)]]] + 2*(a^2 - a*b - b^2)*ArcTanh[(Sqrt[2*a + 2*b]*Sin[e + f*x])/ Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + Sqrt[a + b]*(4*b^(3/2)*Log[Sqrt[2*a + b - b*Cos[2*(e + f*x)]] + Sqrt[2]*Sqrt[b]*Sin[e + f*x]] + Sqrt[2]*(a + b )*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Sec[e + f*x]*Tan[e + f*x]))/(4*Sqrt[a + b]*f)
Time = 0.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3669, 315, 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.
| \(\displaystyle \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\cos (e+f x)^3}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}-\frac {1}{2} \int -\frac {a (a-b)-2 b^2 \sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {a (a-b)-2 b^2 \sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b^2 \int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+(a-2 b) (a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b^2 \int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+(a-2 b) (a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left ((a-2 b) (a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{2} \left ((a-2 b) (a+b) \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )+(a-2 b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )\right )+\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
((2*b^(3/2)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]] + ( a - 2*b)*Sqrt[a + b]*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/2 + ((a + b)*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*(1 - S in[e + f*x]^2)))/f
3.4.37.3.1 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[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 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. \(401\) vs. \(2(109)=218\).
Time = 1.48 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.17
| method | result | size |
| default | \(\frac {-\left (-4 b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) \left (a +b \right )^{\frac {3}{2}}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{3}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{2}-2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}}{4 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2} f}\) | \(402\) |
1/4*(-(-4*b^(3/2)*ln(b^(1/2)*sin(f*x+e)+(a+b-b*cos(f*x+e)^2)^(1/2))*(a+b)^ (3/2)-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f* x+e)+a))*a^3+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2) +b*sin(f*x+e)+a))*a*b^2+2*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+ e)^2)^(1/2)+b*sin(f*x+e)+a))*b^3+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3-3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2) *(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^2-2*ln(2/(1+sin(f*x+e))*( (a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^3)*cos(f*x+e)^2+ 2*sin(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2))/(a+b)^(3/2)/cos(f*x+e )^2/f
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (109) = 218\).
Time = 0.71 (sec) , antiderivative size = 1471, normalized size of antiderivative = 11.58 \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/8*(b^(3/2)*cos(f*x + e)^2*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)*sqr t(b)*sin(f*x + e)) - sqrt(a + b)*(a - 2*b)*cos(f*x + e)^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) + 4*sqrt(-b*c os(f*x + e)^2 + a + b)*(a + b)*sin(f*x + e))/(f*cos(f*x + e)^2), -1/8*(2*( a - 2*b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sq rt(-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)))*cos(f*x + e)^2 - b^(3/2)*cos(f*x + e)^2* 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*sqr...
Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^3} \,d x \]