3.315 \(\int (a+a \cos (c+d x))^4 \sqrt {\sec (c+d x)} \, dx\)

Optimal. Leaf size=161 \[ \frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {136 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {64 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

[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

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Rubi [A]  time = 0.23, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3238, 3791, 3769, 3771, 2641, 2639} \[ \frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {136 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {64 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*Sqrt[Sec[c + d*x]],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 2639

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

Rule 2641

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

Rule 3238

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 3769

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 3771

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 3791

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*} \int (a+a \cos (c+d x))^4 \sqrt {\sec (c+d x)} \, dx &=\int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\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)} F\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)}{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)} F\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)}{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)} F\left (\left .\frac {1}{2} (c+d x)\right |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*}

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Mathematica [C]  time = 1.56, size = 146, normalized size = 0.91 \[ \frac {a^4 \left (\frac {10752 i \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-2720 i \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right ) \sec (c+d x)+1910 \sin (c+d x)+336 \sin (2 (c+d x))+30 \sin (3 (c+d x))-5376 i\right )}{420 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(-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]])

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fricas [F]  time = 1.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*cos(d*x + c)^4 + 4*a^4*cos(d*x + c)^3 + 6*a^4*cos(d*x + c)^2 + 4*a^4*cos(d*x + c) + a^4)*sqrt(se
c(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.54, size = 272, normalized size = 1.69 \[ -\frac {8 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-258 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+448 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-167 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-167*sin(1/2*d*x+1/2*c)^2*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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int 4 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}\, dx + \int 6 \cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}\, dx + \int 4 \cos ^{3}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \cos ^{4}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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(sq
rt(sec(c + d*x)), x))

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