Integrand size = 23, antiderivative size = 70 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \sin (c+d x)}{d}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin ^3(c+d x)}{3 d} \] Output:
sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+1/3*sec(d*x+c)^(5/2)*(b *sec(d*x+c))^(1/2)*sin(d*x+c)^3/d
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.64 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {b \sec (c+d x)} \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d \sqrt {\sec (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^(7/2)*Sqrt[b*Sec[c + d*x]],x]
Output:
(Sqrt[b*Sec[c + d*x]]*(Tan[c + d*x] + Tan[c + d*x]^3/3))/(d*Sqrt[Sec[c + d *x]])
Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2031, 3042, 4254, 2009}
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.
\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \sec ^4(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {\sqrt {b \sec (c+d x)} \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d \sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right ) \sqrt {b \sec (c+d x)}}{d \sqrt {\sec (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^(7/2)*Sqrt[b*Sec[c + d*x]],x]
Output:
-((Sqrt[b*Sec[c + d*x]]*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/(d*Sqrt[Sec[c + d*x]]))
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.76 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {\sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right ) \sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{3 d}\) | \(48\) |
risch | \(\frac {4 i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (4 \cos \left (d x +c \right )+2 i \sin \left (d x +c \right )\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}\) | \(89\) |
Input:
int(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3/d*sin(d*x+c)*(2*cos(d*x+c)^2+1)*sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2)* cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {5}{2}}} \] Input:
integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/3*(2*cos(d*x + c)^2 + 1)*sqrt(b/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^(5/2))
Timed out. \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(7/2)*(b*sec(d*x+c))**(1/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (60) = 120\).
Time = 0.23 (sec) , antiderivative size = 294, normalized size of antiderivative = 4.20 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {4 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{3 \, {\left (2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \] Input:
integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="maxima")
Output:
4/3*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1 )*sin(4*d*x + 4*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4 *c)*sin(2*d*x + 2*c))*sqrt(b)/((2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*c os(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d *x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*si n(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2* c)^2 + 6*cos(2*d*x + 2*c) + 1)*d)
Time = 0.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=-\frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {b} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} d} \] Input:
integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="giac")
Output:
-2/3*(3*tan(1/2*d*x + 1/2*c)^5 - 2*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2*c))*sqrt(b)*sgn(cos(d*x + c))/((tan(1/2*d*x + 1/2*c)^6 - 3*tan(1/2*d *x + 1/2*c)^4 + 3*tan(1/2*d*x + 1/2*c)^2 - 1)*d)
Time = 12.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.80 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (4\,\sin \left (c+d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,10{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,5{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}\right )}{3\,d\,\left (10\,\cos \left (c+d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )\right )} \] Input:
int((b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(7/2),x)
Output:
(2*cos(c + d*x)*(b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)*(cos(c + d*x )*10i + 4*sin(c + d*x) + cos(3*c + 3*d*x)*5i + cos(5*c + 5*d*x)*1i + 5*sin (3*c + 3*d*x) + sin(5*c + 5*d*x)))/(3*d*(10*cos(c + d*x) + 5*cos(3*c + 3*d *x) + cos(5*c + 5*d*x)))
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.64 \[ \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {b}\, \sin \left (d x +c \right ) \left (2 \sin \left (d x +c \right )^{2}-3\right )}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x)
Output:
(sqrt(b)*sin(c + d*x)*(2*sin(c + d*x)**2 - 3))/(3*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))