\(\int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [191]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {64 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]

[Out]

2/7*a^4*sin(d*x+c)/d/sec(d*x+c)^(5/2)+8/5*a^4*sin(d*x+c)/d/sec(d*x+c)^(3/2)+94/21*a^4*sin(d*x+c)/d/sec(d*x+c)^
(1/2)+64/5*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c
)^(1/2)*sec(d*x+c)^(1/2)/d+136/21*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/
2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3876, 3854, 3856, 2720, 2719} \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {136 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {64 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

[In]

Int[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(7/2),x]

[Out]

(64*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (136*a^4*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^4*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (8*a^4*S
in[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (94*a^4*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(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]
 && IGtQ[m, 0] && RationalQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^4}{\sec ^{\frac {7}{2}}(c+d x)}+\frac {4 a^4}{\sec ^{\frac {5}{2}}(c+d x)}+\frac {6 a^4}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^4}{\sqrt {\sec (c+d x)}}+a^4 \sqrt {\sec (c+d x)}\right ) \, dx \\ & = a^4 \int \frac {1}{\sec ^{\frac {7}{2}}(c+d x)} \, dx+a^4 \int \sqrt {\sec (c+d x)} \, dx+\left (4 a^4\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\left (6 a^4\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^4 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {1}{7} \left (5 a^4\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\left (2 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (12 a^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\left (a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (4 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {8 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (5 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\left (2 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (12 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {64 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 {6 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (5 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {64 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 {136 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.93 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^4 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \left (-5376 i+\frac {10752 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-2720 i \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right ) \sec (c+d x)+1910 \sin (c+d x)+336 \sin (2 (c+d x))+30 \sin (3 (c+d x))\right )}{420 d \sqrt {\sec (c+d x)}} \]

[In]

Integrate[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(7/2),x]

[Out]

(a^4*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(-5376*I + ((10752*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -
E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - (2720*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] + 1910*Sin[c + d*x] + 336*Sin[2*(c + d*x)] + 30*Sin[3*(c +
d*x)]))/(420*d*Sqrt[Sec[c + d*x]])

Maple [A] (verified)

Time = 25.63 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.69

method result size
default \(-\frac {8 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-258 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+448 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-167 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(272\)
parts \(\text {Expression too large to display}\) \(858\)

[In]

int((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-8/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(60*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-258*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-167*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^2+85*(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))-168*(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))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^
(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (15 \, a^{4} \cos \left (d x + c\right )^{3} + 84 \, a^{4} \cos \left (d x + c\right )^{2} + 235 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-2/105*(170*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 170*I*sqrt(2)*a^4*weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 336*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 336*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
 0, cos(d*x + c) - I*sin(d*x + c))) - (15*a^4*cos(d*x + c)^3 + 84*a^4*cos(d*x + c)^2 + 235*a^4*cos(d*x + c))*s
in(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=a^{4} \left (\int \frac {1}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {6}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**4/sec(d*x+c)**(7/2),x)

[Out]

a**4*(Integral(sec(c + d*x)**(-7/2), x) + Integral(4/sec(c + d*x)**(5/2), x) + Integral(6/sec(c + d*x)**(3/2),
 x) + Integral(4/sqrt(sec(c + d*x)), x) + Integral(sqrt(sec(c + d*x)), x))

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(7/2), x)

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

[In]

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(7/2),x)

[Out]

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(7/2), x)