\(\int \frac {(a+a \cos (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx\) [316]

   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 = 187 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx=\frac {152 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{7 d}+\frac {2 a^4 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {122 a^4 \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {32 a^4 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}} \]

[Out]

2/9*a^4*sin(d*x+c)/d/sec(d*x+c)^(7/2)+8/7*a^4*sin(d*x+c)/d/sec(d*x+c)^(5/2)+122/45*a^4*sin(d*x+c)/d/sec(d*x+c)
^(3/2)+32/7*a^4*sin(d*x+c)/d/sec(d*x+c)^(1/2)+152/15*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+32/7*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 = 0.33 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3317, 3876, 3854, 3856, 2719, 2720} \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx=\frac {122 a^4 \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {32 a^4 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {32 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d}+\frac {152 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d} \]

[In]

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

[Out]

(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) + (2*a^4*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (8*a^4*S
in[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (122*a^4*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (32*a^4*Sin[c + d*x
])/(7*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 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

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 \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \int \left (\frac {a^4}{\sec ^{\frac {9}{2}}(c+d x)}+\frac {4 a^4}{\sec ^{\frac {7}{2}}(c+d x)}+\frac {6 a^4}{\sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^4}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {a^4}{\sqrt {\sec (c+d x)}}\right ) \, dx \\ & = a^4 \int \frac {1}{\sec ^{\frac {9}{2}}(c+d x)} \, dx+a^4 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\left (4 a^4\right ) \int \frac {1}{\sec ^{\frac {7}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\left (6 a^4\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^4 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {12 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 {1}{9} \left (7 a^4\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{3} \left (4 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{7} \left (20 a^4\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{5} \left (18 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 \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 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 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {122 a^4 \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {32 a^4 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{15} \left (7 a^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (20 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{3} \left (4 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 (18 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {46 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 {8 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)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {122 a^4 \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {32 a^4 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{15} \left (7 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (20 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, 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 {2 a^4 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {122 a^4 \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {32 a^4 \sin (c+d x)}{7 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 = 3.24 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.83 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^4 \left (-25536 i+\frac {51072 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-11520 i \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right ) \sec (c+d x)+12240 \sin (c+d x)+3556 \sin (2 (c+d x))+720 \sin (3 (c+d x))+70 \sin (4 (c+d x))\right )}{2520 d \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

(a^4*(-25536*I + ((51072*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*
x))] - (11520*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] + 12240*Sin[c + d*x] + 3556*Sin[2*(c + d*x)] + 720*Sin[3*(c + d*x)] + 70*Sin[4*(c + d*x)]))/(2520*d*Sqrt[
Sec[c + d*x]])

Maple [A] (verified)

Time = 16.32 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.39

method result size
default \(-\frac {8 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{4} \left (280 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+34 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+72 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-485 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+180 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-399 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+219 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(260\)
parts \(\text {Expression too large to display}\) \(937\)

[In]

int((a+cos(d*x+c)*a)^4/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-8/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(280*cos(1/2*d*x+1/2*c)^11-120*cos(1/2*d*x+
1/2*c)^9+34*cos(1/2*d*x+1/2*c)^7+72*cos(1/2*d*x+1/2*c)^5-485*cos(1/2*d*x+1/2*c)^3+180*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-399*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+219*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)/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 = 0.10 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.98 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (360 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 360 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 798 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 798 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (35 \, a^{4} \cos \left (d x + c\right )^{4} + 180 \, a^{4} \cos \left (d x + c\right )^{3} + 427 \, a^{4} \cos \left (d x + c\right )^{2} + 720 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-2/315*(360*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 360*I*sqrt(2)*a^4*weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 798*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 798*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
 0, cos(d*x + c) - I*sin(d*x + c))) - (35*a^4*cos(d*x + c)^4 + 180*a^4*cos(d*x + c)^3 + 427*a^4*cos(d*x + c)^2
 + 720*a^4*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F]

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

[In]

integrate((a+a*cos(d*x+c))**4/sec(d*x+c)**(1/2),x)

[Out]

a**4*(Integral(4*cos(c + d*x)/sqrt(sec(c + d*x)), x) + Integral(6*cos(c + d*x)**2/sqrt(sec(c + d*x)), x) + Int
egral(4*cos(c + d*x)**3/sqrt(sec(c + d*x)), x) + Integral(cos(c + d*x)**4/sqrt(sec(c + d*x)), x) + Integral(1/
sqrt(sec(c + d*x)), x))

Maxima [F]

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

[In]

integrate((a+a*cos(d*x+c))^4/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)

Giac [F]

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

[In]

integrate((a+a*cos(d*x+c))^4/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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