Integrand size = 23, antiderivative size = 159 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {56 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:
56/5*a^4*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c) ^(1/2)/d+32/3*a^4*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))* sec(d*x+c)^(1/2)/d+2/5*a^4*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/3*a^4*sin(d*x+c )/d/sec(d*x+c)^(1/2)+2*a^4*sec(d*x+c)^(1/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^4 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \left (-336 i+\frac {672 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-320 i \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right ) \sec (c+d x)+80 \sin (c+d x)+3 \sec (c+d x) \sin (3 (c+d x))+63 \tan (c+d x)\right )}{30 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]
Output:
(a^4*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(-336*I + ((672*I)*Hy pergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*( c + d*x))] - (320*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]*Sec[c + d*x] + 80*Sin[c + d*x] + 3*Sec[c + d*x]*Sin[3*(c + d*x)] + 63*Tan[c + d*x]))/(30*d*Sqrt[Sec[c + d*x]])
Time = 0.40 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4278, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \int \left (a^4 \sec ^{\frac {3}{2}}(c+d x)+\frac {4 a^4}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {a^4}{\sec ^{\frac {5}{2}}(c+d x)}+4 a^4 \sqrt {\sec (c+d x)}+\frac {6 a^4}{\sqrt {\sec (c+d x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {32 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {56 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]
Output:
(56*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/( 5*d) + (32*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d *x]])/(3*d) + (2*a^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^4*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x] )/d
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Time = 5.97 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(\frac {8 a^{4} \left (6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-26 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+19 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(194\) |
| parts | \(\text {Expression too large to display}\) | \(841\) |
Input:
int((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
8/15*a^4*(6*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-26*sin(1/2*d*x+1/2*c)^ 4*cos(1/2*d*x+1/2*c)+19*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-20*(sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1 /2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos (1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (40 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 40 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 42 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, a^{4} \cos \left (d x + c\right )^{2} + 20 \, a^{4} \cos \left (d x + c\right ) + 15 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \] Input:
integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="fricas")
Output:
-2/15*(40*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d* x + c)) - 40*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin (d*x + c)) - 42*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c))) + 42*I*sqrt(2)*a^4*weierstrassZeta(- 4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (3*a^4* cos(d*x + c)^2 + 20*a^4*cos(d*x + c) + 15*a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=a^{4} \left (\int \frac {1}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {6}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 4 \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**4/sec(d*x+c)**(5/2),x)
Output:
a**4*(Integral(sec(c + d*x)**(-5/2), x) + Integral(4/sec(c + d*x)**(3/2), x) + Integral(6/sqrt(sec(c + d*x)), x) + Integral(4*sqrt(sec(c + d*x)), x) + Integral(sec(c + d*x)**(3/2), x))
\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)
\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(5/2),x)
Output:
int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(5/2), x)
\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=a^{4} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x +4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right )+6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right )+4 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) \] Input:
int((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x)
Output:
a**4*(int(sqrt(sec(c + d*x))/sec(c + d*x)**3,x) + 4*int(sqrt(sec(c + d*x)) /sec(c + d*x)**2,x) + 6*int(sqrt(sec(c + d*x))/sec(c + d*x),x) + 4*int(sqr t(sec(c + d*x)),x) + int(sqrt(sec(c + d*x))*sec(c + d*x),x))