Integrand size = 33, antiderivative size = 240 \[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=-\frac {2 \sec ^{\frac {3}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{\left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}-\frac {2 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sec ^{\frac {3}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^2}{\left (a^2-b^2+c^2\right ) e \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \]
-2*sec(e*x+d)^(3/2)*(c*cos(e*x+d)-a*sin(e*x+d))*(b+a*cos(e*x+d)+c*sin(e*x+ d))/(a^2-b^2+c^2)/e/(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)-2*(cos(1/2*d+1/2*e *x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))*EllipticE( sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^( 1/2)))^(1/2))*sec(e*x+d)^(3/2)*(b+a*cos(e*x+d)+c*sin(e*x+d))^2/(a^2-b^2+c^ 2)/e/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)/(a+b*sec(e* x+d)+c*tan(e*x+d))^(3/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 6.94 (sec) , antiderivative size = 1732, normalized size of antiderivative = 7.22 \[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx =\text {Too large to display} \]
(Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2*((-2*(a^2 + c^ 2))/(a*c*(a^2 - b^2 + c^2)) + (2*(b*c + a^2*Sin[d + e*x] + c^2*Sin[d + e*x ]))/(a*(a^2 - b^2 + c^2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x]))))/(e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (2*b*AppellF1[1/2, 1/2, 1/2, 3/2 , -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2 ]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e* x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*S ec[d + e*x]^(3/2)*Sec[d + e*x + ArcTan[a/c]]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin [d + e*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a ^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c ^2)/c^2])])/(Sqrt[1 + a^2/c^2]*c*(a^2 - b^2 + c^2)*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (a^2*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c* Sin[d + e*x])^(3/2)*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(1 - b/(a*Sqrt [1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/ (a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d + e*x - ArcTa n[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c^2)/a^2])]*...
Time = 0.74 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 3646, 3042, 3607, 3042, 3598, 3042, 3132}
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.
\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (d+e x)^{3/2}}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}dx\) |
\(\Big \downarrow \) 3646 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \int \frac {1}{(b+a \cos (d+e x)+c \sin (d+e x))^{3/2}}dx}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \int \frac {1}{(b+a \cos (d+e x)+c \sin (d+e x))^{3/2}}dx}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3607 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \left (-\frac {\int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{a^2-b^2+c^2}-\frac {2 (c \cos (d+e x)-a \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\right )}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \left (-\frac {\int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{a^2-b^2+c^2}-\frac {2 (c \cos (d+e x)-a \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\right )}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3598 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \left (-\frac {\sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}dx}{\left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}-\frac {2 (c \cos (d+e x)-a \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\right )}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \left (-\frac {\sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}dx}{\left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}-\frac {2 (c \cos (d+e x)-a \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\right )}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2} \left (-\frac {2 \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}-\frac {2 (c \cos (d+e x)-a \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\right )}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}\) |
(Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)*((-2*(c*Co s[d + e*x] - a*Sin[d + e*x]))/((a^2 - b^2 + c^2)*e*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]) - (2*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]) /((a^2 - b^2 + c^2)*e*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt [a^2 + c^2])])))/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)
3.5.51.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ )]], x_Symbol] :> Simp[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])] Int[Sqrt[a/(a + S qrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - Arc Tan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-3/2), x_Symbol] :> Simp[2*((c*Cos[d + e*x] - b*Sin[d + e*x])/(e*(a^2 - b^ 2 - c^2)*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])), x] + Simp[1/(a^2 - b^ 2 - c^2) Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{ a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_), x_Symbol] :> Simp[Sec[d + e*x]^n*((b + a*Cos[d + e*x] + c*Sin[d + e*x])^n/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n ) Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c , d, e}, x] && EqQ[m + n, 0] && !IntegerQ[n]
Result contains complex when optimal does not.
Time = 20.45 (sec) , antiderivative size = 40064, normalized size of antiderivative = 166.93
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 1724, normalized size of antiderivative = 7.18 \[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=\text {Too large to display} \]
1/3*((I*a*b^2 - b^2*c + (I*a^2*b - a*b*c)*cos(e*x + d) + (I*a*b*c - b*c^2) *sin(e*x + d))*sqrt(2*a - 2*I*c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b ^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2 *c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9*a ^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2*b^3)* c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b + 2*I*b*c + 3*(a^2 + c^ 2)*cos(e*x + d) - 3*(-I*a^2 - I*c^2)*sin(e*x + d))/(a^2 + c^2)) + (-I*a*b^ 2 - b^2*c + (-I*a^2*b - a*b*c)*cos(e*x + d) + (-I*a*b*c - b*c^2)*sin(e*x + d))*sqrt(2*a + 2*I*c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2 *c^2 - 6*I*a*c^3 - 3*c^4 - 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c^4 ), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 + 9*I*b*c^5 - 2*I*(9*a^2*b + 4*b ^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 - 3*I*(9*a^4*b - 8*a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b - 2*I*b*c + 3*(a^2 + c^2)*cos(e*x + d) - 3*(I*a^2 + I*c^2)*sin(e*x + d))/(a^2 + c^2)) - 3*(I*a^2*b + I*b*c^ 2 + (I*a^3 + I*a*c^2)*cos(e*x + d) + (I*a^2*c + I*c^3)*sin(e*x + d))*sqrt( 2*a - 2*I*c)*weierstrassZeta(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c ^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5 *b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9*a^2*b + 4*b^3)*c^3 - 6*(3 *a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3* a^2*c^4 + c^6), weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2*c^2...
\[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=\int \frac {\sec ^{\frac {3}{2}}{\left (d + e x \right )}}{\left (a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=\int { \frac {\sec \left (e x + d\right )^{\frac {3}{2}}}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (d+e\,x\right )}\right )}^{3/2}}{{\left (a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}\right )}^{3/2}} \,d x \]