3.172 \(\int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {56 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {66 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-56/5*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))/d+32/3*a^4*(co
s(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))/d+2/5*a^4*sin(d*x+c)/d/cos(
d*x+c)^(5/2)+8/3*a^4*sin(d*x+c)/d/cos(d*x+c)^(3/2)+66/5*a^4*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2757, 2636, 2639, 2641} \[ \frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {56 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {66 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-56*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (32*a^4*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^4*Sin[c + d*x])/(5
*d*Cos[c + d*x]^(5/2)) + (8*a^4*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (66*a^4*Sin[c + d*x])/(5*d*Sqrt[Cos[c
+ d*x]])

Rule 2636

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

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 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\int \left (\frac {a^4}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a^4}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {6 a^4}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^4}{\sqrt {\cos (c+d x)}}+a^4 \sqrt {\cos (c+d x)}\right ) \, dx\\ &=a^4 \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)} \, dx+a^4 \int \sqrt {\cos (c+d x)} \, dx+\left (4 a^4\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (6 a^4\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {12 a^4 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{5} \left (3 a^4\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{3} \left (4 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\left (6 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {10 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {66 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} \left (3 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {56 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {66 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 4.34, size = 283, normalized size = 2.34 \[ \frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {168 \sec (c) \cos ^2(c+d x) \left (\csc (c) \sqrt {\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \left (3 \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+\cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-2 \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )\right )}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-640 \sin (c) \sqrt {\csc ^2(c)} \cos ^3(c+d x) \sqrt {\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\csc (c) (141 \cos (2 c+d x)+40 \cos (c+2 d x)-40 \cos (3 c+2 d x)+183 \cos (2 c+3 d x)-15 \cos (4 c+3 d x)+363 \cos (d x))\right )}{960 d \cos ^{\frac {5}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*((363*Cos[d*x] + 141*Cos[2*c + d*x] + 40*Cos[c + 2*d*x] - 40*Cos[
3*c + 2*d*x] + 183*Cos[2*c + 3*d*x] - 15*Cos[4*c + 3*d*x])*Csc[c] - 640*Cos[c + d*x]^3*Sqrt[Cos[d*x - ArcTan[C
ot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[C
ot[c]]]*Sin[c] - (168*Cos[c + d*x]^2*Sec[c]*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]
]]^2]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Csc[c]*Sqr
t[Sin[d*x + ArcTan[Tan[c]]]^2]))/(Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])))/(960*d*Cos[c + d*x]^(5/2
))

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fricas [F]  time = 1.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.90, size = 386, normalized size = 3.19 \[ -\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 {7 \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 (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{20 \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 {41 \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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{60 \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 )}}{320 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {33 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{40 \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 {\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 )}}{24 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{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/cos(d*x+c)^(7/2),x)

[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*(-7/20*(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)))+41/60*(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))-1/320*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^3-33/40*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)-1/24*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2)/sin(1/2*d*x+1/2*c)/(2*c
os(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 \[ \int \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.30, size = 202, normalized size = 1.67 \[ \frac {2\,\left (a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,\left (\frac {34\,a^4\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {a^4\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d}+\frac {8\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a^4*ellipticE(c/2 + (d*x)/2, 2) + 4*a^4*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*((34*a^4*sin(c + d*x))/(cos(c
+ d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (a^4*sin(c + d*x))/(cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)))*hyperg
eom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(5*d) + (8*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2)
)/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) - (8*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 7/4, cos(c + d*
x)^2))/(15*d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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