Integrand size = 35, antiderivative size = 124 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^{3/2} (2 A+3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^2 (2 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d} \]
a^(3/2)*(2*A+3*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+a^2 *(2*A-B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+a*B*sin(d*x+ c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
Time = 1.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 \left (2 A \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-3 B \arcsin \left (\sqrt {\sec (c+d x)}\right )+(B+2 A \cos (c+d x)) \sqrt {(-1+\cos (c+d x)) \sec ^2(c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
(a^2*(2*A*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 3*B*ArcSin[Sqrt[Sec[c + d*x]]] + (B + 2*A*Cos[c + d*x])*Sqrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.72 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4506, 27, 3042, 4503, 3042, 4288, 222}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x) a+a} (a (2 A-B)+a (2 A+3 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} (a (2 A-B)+a (2 A+3 B) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (2 A-B)+a (2 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{2} \left (a (2 A+3 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 (2 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a (2 A+3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (2 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a^2 (2 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (2 A+3 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a^{3/2} (2 A+3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (2 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\) |
(a*B*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((2*a^( 3/2)*(2*A + 3*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]) /d + (2*a^2*(2*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2
3.3.34.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(334\) vs. \(2(110)=220\).
Time = 8.02 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.70
| method | result | size |
| parts | \(-\frac {A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}+\frac {B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (3 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(335\) |
| default | \(\frac {a \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+2 A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+3 B \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+3 B \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-4 A \cot \left (d x +c \right ) \cos \left (d x +c \right )^{2}+4 A \cot \left (d x +c \right ) \cos \left (d x +c \right )-2 B \cos \left (d x +c \right ) \cot \left (d x +c \right )+2 B \cot \left (d x +c \right )\right )}{2 d}\) | \(352\) |
-A/d*a*(a*(1+sec(d*x+c)))^(1/2)/sec(d*x+c)^(1/2)*((-1/(cos(d*x+c)+1))^(1/2 )*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^ (1/2))-(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos (d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+2*cot(d*x+c)-2*csc(d*x+c))+1/2*B/d*a *(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(1/2)*(3*cos(d*x+c)*arctan(1/2*(cos(d *x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-3*cos(d*x+c) *arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^( 1/2))+2*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2))/(cos(d*x+c)+1)/(-1/(cos(d*x+ c)+1))^(1/2)
Time = 0.35 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.94 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\left [\frac {{\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + {\left (2 \, A + 3 \, B\right )} a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {{\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + {\left (2 \, A + 3 \, B\right )} a\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/4*(((2*A + 3*B)*a*cos(d*x + c) + (2*A + 3*B)*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)* sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(2*A*a*cos(d*x + c) + B*a)*sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*c os(d*x + c) + d), 1/2*(((2*A + 3*B)*a*cos(d*x + c) + (2*A + 3*B)*a)*sqrt(- a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(2*A*a*c os(d*x + c) + B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sq rt(cos(d*x + c)))/(d*cos(d*x + c) + d)]
\[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right )}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1417 vs. \(2 (110) = 220\).
Time = 0.58 (sec) , antiderivative size = 1417, normalized size of antiderivative = 11.43 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \]
1/4*(sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2 *c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2 *sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt (2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)* cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log (2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d *x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1/2* c))*A*sqrt(a) + (3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c )^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*c os(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/ 2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2* c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*si n(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2 *c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt (2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2 *d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^...
\[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]