Integrand size = 23, antiderivative size = 126 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{5/2} f}+\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b)^(5/2)/f+1/3 *b*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)+1/3*b*(5*a+2*b)*sin(f*x+e )/a^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.57 (sec) , antiderivative size = 1291, normalized size of antiderivative = 10.25 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]
(Sec[e + f*x]*Tan[e + f*x]*(1575*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a) ]] + (2100*b*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2)/a + (840*b^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^4)/a^ 2 + (3150*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Tan[e + f*x] ^2)/a + (4200*b*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2*Tan[e + f*x]^2)/a^2 + (1680*b^2*(a + b)*ArcSin[Sqrt[-(((a + b)*Ta n[e + f*x]^2)/a)]]*Sin[e + f*x]^4*Tan[e + f*x]^2)/a^3 + (1575*(a + b)^2*Ar cSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Tan[e + f*x]^4)/a^2 + (2100*b*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2*Tan[e + f*x]^4)/a^3 + (840*b^2*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a )]]*Sin[e + f*x]^4*Tan[e + f*x]^4)/a^4 + 2100*Sqrt[(Sec[e + f*x]^2*(a + b* Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) + (2800*b*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))/a + (1120*b^2*Sin[e + f*x]^4*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))/a^2 + 96*H ypergeometric2F1[2, 2, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f *x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 2 4*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Tan[e + f*x]^2)/a)]*Sq rt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/ a))^(7/2) + (168*b*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Tan[e + f*x]...
Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3669, 316, 402, 27, 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.
\(\displaystyle \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (e+f x) \left (a+b \sin (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\frac {b \sin (e+f x)}{3 a (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\int \frac {2 b \sin ^2(e+f x)+b-3 (a+b)}{\left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a (a+b)}}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {b \sin (e+f x)}{3 a (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {-\frac {\int \frac {3 a^2}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a (a+b)}-\frac {b (5 a+2 b) \sin (e+f x)}{a (a+b) \sqrt {a+b \sin ^2(e+f x)}}}{3 a (a+b)}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {b \sin (e+f x)}{3 a (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {-\frac {3 a \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a+b}-\frac {b (5 a+2 b) \sin (e+f x)}{a (a+b) \sqrt {a+b \sin ^2(e+f x)}}}{3 a (a+b)}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {b \sin (e+f x)}{3 a (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {-\frac {3 a \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}}{a+b}-\frac {b (5 a+2 b) \sin (e+f x)}{a (a+b) \sqrt {a+b \sin ^2(e+f x)}}}{3 a (a+b)}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {b \sin (e+f x)}{3 a (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {-\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{3/2}}-\frac {b (5 a+2 b) \sin (e+f x)}{a (a+b) \sqrt {a+b \sin ^2(e+f x)}}}{3 a (a+b)}}{f}\) |
((b*Sin[e + f*x])/(3*a*(a + b)*(a + b*Sin[e + f*x]^2)^(3/2)) - ((-3*a*ArcT anh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(a + b)^(3/2) - (b*(5*a + 2*b)*Sin[e + f*x])/(a*(a + b)*Sqrt[a + b*Sin[e + f*x]^2]))/(3* a*(a + b)))/f
3.4.66.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]*((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[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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. \(898\) vs. \(2(112)=224\).
Time = 1.63 (sec) , antiderivative size = 899, normalized size of antiderivative = 7.13
method | result | size |
default | \(\frac {-3 a^{4} b^{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 )+3 a^{4} b^{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 )-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^{2} b^{4}+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^{2} b^{4}-6 a^{3} 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 )+6 a^{3} 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}{\sin \left (f x +e \right )-1}\right )+3 a^{2} b^{4} \left (\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 )-\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 )\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}\, b^{4} \left (5 a +2 b \right )+4 \sin \left (f x +e \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}\, b^{3} \left (3 a^{2}+4 a b +b^{2}\right )-6 a^{2} b^{3} \left (\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 +\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 -\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 -\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 \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{6 b^{2} \sqrt {a +b}\, a^{2} \left (\left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b^{2}+2 a \,b^{3} \left (\cos ^{4}\left (f x +e \right )\right )+b^{4} \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b -6 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}-6 a \,b^{3} \left (\cos ^{2}\left (f x +e \right )\right )-2 b^{4} \left (\cos ^{2}\left (f x +e \right )\right )+a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) f}\) | \(899\) |
1/6/b^2/(a+b)^(1/2)/a^2/(cos(f*x+e)^4*a^2*b^2+2*a*b^3*cos(f*x+e)^4+b^4*cos (f*x+e)^4-2*cos(f*x+e)^2*a^3*b-6*cos(f*x+e)^2*a^2*b^2-6*a*b^3*cos(f*x+e)^2 -2*b^4*cos(f*x+e)^2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(-3*a^4*b^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))+3*a^ 4*b^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))-3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*s in(f*x+e)+a))*a^2*b^4+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^2*b^4-6*a^3*b^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))+6*a^3*b^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))+3*a^2*b^4*(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)) -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)))*cos(f*x+e)^4-2*cos(f*x+e)^2*sin(f*x+e)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b ^2)^(1/2)*(a+b)^(1/2)*b^4*(5*a+2*b)+4*sin(f*x+e)*(-b*cos(f*x+e)^2+(a*b^2+b ^3)/b^2)^(1/2)*(a+b)^(1/2)*b^3*(3*a^2+4*a*b+b^2)-6*a^2*b^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+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-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 -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)*cos(f*x+e)^2)/f
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (112) = 224\).
Time = 0.59 (sec) , antiderivative size = 775, normalized size of antiderivative = 6.15 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left (6 \, a^{3} b + 14 \, a^{2} b^{2} + 10 \, a b^{3} + 2 \, b^{4} - {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{12 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) - 2 \, {\left (6 \, a^{3} b + 14 \, a^{2} b^{2} + 10 \, a b^{3} + 2 \, b^{4} - {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \]
[1/12*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^ 2*b^2)*cos(f*x + e)^2)*sqrt(a + b)*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*(6*a^3*b + 14*a^2*b^2 + 10*a*b^3 + 2*b^4 - (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)* f*cos(f*x + e)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5) *f*cos(f*x + e)^2 + (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f), -1/6*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*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))) - 2*(6*a^3*b + 14*a^2*b^2 + 10*a*b^3 + 2*b^4 - (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^7 + 5*a^6*b + 10*a^5*b^2 + 1 0*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f)]
\[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (112) = 224\).
Time = 0.47 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.16 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {2 \, b \sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a b} + \frac {6 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3} + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} b + \sqrt {b \sin \left (f x + e\right )^{2} + a} a b^{2}} + \frac {4 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3} + \sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} b} + \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}} + \frac {3 \, \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}}}{6 \, f} \]
1/6*(2*b*sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*a^2 + (b*sin(f*x + e)^ 2 + a)^(3/2)*a*b) + 6*b*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^3 + 2*s qrt(b*sin(f*x + e)^2 + a)*a^2*b + sqrt(b*sin(f*x + e)^2 + a)*a*b^2) + 4*b* sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^3 + sqrt(b*sin(f*x + e)^2 + a)* a^2*b) + 3*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)^(5/2) + 3*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)^(5/2 ))/f
\[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]