Integrand size = 23, antiderivative size = 213 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=-\frac {152 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{7 d}+\frac {152 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {32 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {122 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {8 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d} \] Output:
-152/15*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+32/7*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+152/15*a^4*sec(d*x+c)^(1/2)*sin(d*x+c)/d+32/7*a^4*se c(d*x+c)^(3/2)*sin(d*x+c)/d+122/45*a^4*sec(d*x+c)^(5/2)*sin(d*x+c)/d+8/7*a ^4*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*a^4*sec(d*x+c)^(9/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.01 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.36 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {a^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 \left (-\frac {12 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \left (133 \left (1+e^{2 i (c+d x)}\right )+133 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+60 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {1596 \cos (d x) \csc (c)+\left (720+427 \sec (c+d x)+180 \sec ^2(c+d x)+35 \sec ^3(c+d x)\right ) \tan (c+d x)}{\sec ^{\frac {7}{2}}(c+d x)}\right )}{2520 d} \] Input:
Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4,x]
Output:
(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(((-12*I)*Sqrt[2]*Sqrt[E^(I*( c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Cos[c + d*x]^4*(133*(1 + E^((2*I)*(c + d*x))) + 133*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeome tric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 60*E^(I*(c + d*x))*(-1 + E ^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (1596*Cos[ d*x]*Csc[c] + (720 + 427*Sec[c + d*x] + 180*Sec[c + d*x]^2 + 35*Sec[c + d* x]^3)*Tan[c + d*x])/Sec[c + d*x]^(7/2)))/(2520*d)
Time = 0.50 (sec) , antiderivative size = 213, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4dx\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \int \left (a^4 \sec ^{\frac {11}{2}}(c+d x)+4 a^4 \sec ^{\frac {9}{2}}(c+d x)+6 a^4 \sec ^{\frac {7}{2}}(c+d x)+4 a^4 \sec ^{\frac {5}{2}}(c+d x)+a^4 \sec ^{\frac {3}{2}}(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^4 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{9 d}+\frac {8 a^4 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {122 a^4 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{45 d}+\frac {32 a^4 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{7 d}+\frac {152 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d}-\frac {152 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}\) |
Input:
Int[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4,x]
Output:
(-152*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]) /(15*d) + (32*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (152*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (32*a^ 4*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(7*d) + (122*a^4*Sec[c + d*x]^(5/2)*Sin [c + d*x])/(45*d) + (8*a^4*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d) + (2*a^4 *Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*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]
Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(184)=368\).
Time = 8.24 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.31
| method | result | size |
| default | \(-\frac {a^{4} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{72 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{5}}-\frac {61 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{90 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}-\frac {304 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {1544 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {152 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{7 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{4}}-\frac {16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{7 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}\right )}{\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}\) | \(492\) |
| parts | \(\text {Expression too large to display}\) | \(1391\) |
Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-a^4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/72*cos( 1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1 /2*d*x+1/2*c)^2-1/2)^5-61/90*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-304/15*sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2 *c)^2)^(1/2)+1544/105*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^ 2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))-152/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/ 2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^( 1/2)))-1/7*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-16/7*cos(1/2*d*x+1/2*c)*(-2*sin(1/2* d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^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 = 228, normalized size of antiderivative = 1.07 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=-\frac {2 \, {\left (360 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 360 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 798 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 798 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (1596 \, a^{4} \cos \left (d x + c\right )^{4} + 720 \, a^{4} \cos \left (d x + c\right )^{3} + 427 \, a^{4} \cos \left (d x + c\right )^{2} + 180 \, a^{4} \cos \left (d x + c\right ) + 35 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="fricas")
Output:
-2/315*(360*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c)) - 360*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 798*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c))) - 798*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (1596*a^4*cos(d* x + c)^4 + 720*a^4*cos(d*x + c)^3 + 427*a^4*cos(d*x + c)^2 + 180*a^4*cos(d *x + c) + 35*a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**4,x)
Output:
Timed out
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="maxima")
Output:
Timed out
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int((a + a/cos(c + d*x))^4*(1/cos(c + d*x))^(3/2),x)
Output:
int((a + a/cos(c + d*x))^4*(1/cos(c + d*x))^(3/2), x)
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx=a^{4} \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}d x +4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right )+6 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right )+4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) \] Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x)
Output:
a**4*(int(sqrt(sec(c + d*x))*sec(c + d*x)**5,x) + 4*int(sqrt(sec(c + d*x)) *sec(c + d*x)**4,x) + 6*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) + int(sqrt(sec(c + d*x))*sec(c + d*x ),x))