\(\int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx\) [311]

Optimal result

Integrand size = 23, antiderivative size = 187 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {64 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {64 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {94 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

94/21*a^4*sec(d*x+c)^(3/2)*sin(d*x+c)/d+8/5*a^4*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*a^4*sec(d*x+c)^(7/2)*sin(d*x
+c)/d+64/5*a^4*sin(d*x+c)*sec(d*x+c)^(1/2)/d-64/5*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
E(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 = 0.33 (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, 3856, 2720, 3853, 2719} \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a^4 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {64 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {64 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

[In]

Int[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(9/2),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]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (64*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (94*a^
4*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (8*a^4*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a^4*Sec[c + d*x
]^(7/2)*Sin[c + d*x])/(7*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx \\ & = \int \left (a^4 \sqrt {\sec (c+d x)}+4 a^4 \sec ^{\frac {3}{2}}(c+d x)+6 a^4 \sec ^{\frac {5}{2}}(c+d x)+4 a^4 \sec ^{\frac {7}{2}}(c+d x)+a^4 \sec ^{\frac {9}{2}}(c+d x)\right ) \, dx \\ & = a^4 \int \sqrt {\sec (c+d x)} \, dx+a^4 \int \sec ^{\frac {9}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^{\frac {7}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {8 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {4 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (5 a^4\right ) \int \sec ^{\frac {5}{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 \sec ^{\frac {3}{2}}(c+d x) \, dx-\left (4 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 \\ & = \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 {64 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {94 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 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 (2 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 {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 {64 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {94 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\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 {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 {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 {64 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {94 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.67 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (168 \left (1+e^{2 i (c+d x)}\right )+168 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+85 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\sqrt {\sec (c+d x)} \left (672 \cos (d x) \csc (c)+\left (235+84 \sec (c+d x)+15 \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{840 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(9/2),x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*
(168*(1 + E^((2*I)*(c + d*x))) + 168*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4,
1/2, 3/4, -E^((2*I)*(c + d*x))] + 85*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeo
metric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + Sqrt[Sec[c + d*x]]*(67
2*Cos[d*x]*Csc[c] + (235 + 84*Sec[c + d*x] + 15*Sec[c + d*x]^2)*Tan[c + d*x])))/(840*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(438\) vs. \(2(211)=422\).

Time = 202.27 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.35

method	result	size
default	\(-\frac {32 \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 (\frac {253 \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 )}{420 \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 )}}-\frac {47 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{672 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}-\frac {4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \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 )}}-\frac {2 \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}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \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 )}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{80 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{896 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{4}}\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}\)	\(439\)
parts	\(\text {Expression too large to display}\)	\(1141\)

[In]

int((a+cos(d*x+c)*a)^4*sec(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-32*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(253/420*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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))-47/672*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)^2-4/5*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)-2
/5*(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)))-1/80*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)^3-1/896*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)/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 = 215, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (672 \, a^{4} \cos \left (d x + c\right )^{3} + 235 \, a^{4} \cos \left (d x + c\right )^{2} + 84 \, a^{4} \cos \left (d x + c\right ) + 15 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(170*I*sqrt(2)*a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 170*I*sqr
t(2)*a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 336*I*sqrt(2)*a^4*cos(d*x
+ c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 336*I*sqrt(2)*a^4*c
os(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (672*a^4*cos
(d*x + c)^3 + 235*a^4*cos(d*x + c)^2 + 84*a^4*cos(d*x + c) + 15*a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d
*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**4*sec(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(9/2), x)

Giac [F]

\[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^4 \,d x \]

[In]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^4,x)

[Out]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^4, x)