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

Optimal. Leaf size=121 \[ \frac {136 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

[Out]

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))/d+136/21*a^4*(c
os(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+8/5*a^4*cos(d*x+c)^(3/2)
*sin(d*x+c)/d+2/7*a^4*cos(d*x+c)^(5/2)*sin(d*x+c)/d+94/21*a^4*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.14, 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, 2641, 2639, 2635} \[ \frac {136 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[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}{\sqrt {\cos (c+d x)}} \, dx &=\int \left (\frac {a^4}{\sqrt {\cos (c+d x)}}+4 a^4 \sqrt {\cos (c+d x)}+6 a^4 \cos ^{\frac {3}{2}}(c+d x)+4 a^4 \cos ^{\frac {5}{2}}(c+d x)+a^4 \cos ^{\frac {7}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+a^4 \int \cos ^{\frac {7}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx+\left (4 a^4\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {8 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {4 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (5 a^4\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\left (2 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (12 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {6 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {136 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [C]  time = 6.17, size = 500, normalized size = 4.13 \[ -\frac {2 \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )}{\sqrt {\tan ^2(c)+1} \sqrt {1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt {\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac {\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt {\tan ^2(c)+1}}+\frac {2 \cos ^2(c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{5 d}-\frac {17 \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin (c) \left (-\sqrt {\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{42 d \sqrt {\cot ^2(c)+1}}+\sqrt {\cos (c+d x)} \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac {191 \sin (c) \cos (d x)}{672 d}+\frac {\sin (2 c) \cos (2 d x)}{20 d}+\frac {\sin (3 c) \cos (3 d x)}{224 d}+\frac {191 \cos (c) \sin (d x)}{672 d}+\frac {\cos (2 c) \sin (2 d x)}{20 d}+\frac {\cos (3 c) \sin (3 d x)}{224 d}-\frac {4 \cot (c)}{5 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*((-4*Cot[c])/(5*d) + (191*Cos[d*x]*Sin[c])/(672
*d) + (Cos[2*d*x]*Sin[2*c])/(20*d) + (Cos[3*d*x]*Sin[3*c])/(224*d) + (191*Cos[c]*Sin[d*x])/(672*d) + (Cos[2*c]
*Sin[2*d*x])/(20*d) + (Cos[3*c]*Sin[3*d*x])/(224*d)) - (17*(a + a*Cos[c + d*x])^4*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x -
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c
]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (2*(a + a*Cos[c + d*x])^4*Csc[c]*Sec[c/2 + (d*x)/2]^8*((HypergeometricPFQ[{-
1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[T
an[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1
+ Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*
Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d)

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}}{\sqrt {\cos \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4/cos(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(co
s(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.49, size = 272, normalized size = 2.25 \[ -\frac {8 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-258 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+448 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-167 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.71, size = 146, normalized size = 1.21 \[ \frac {2\,\left (4\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+2\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}-\frac {8\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^4\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(4*a^4*ellipticE(c/2 + (d*x)/2, 2) + 3*a^4*ellipticF(c/2 + (d*x)/2, 2) + 2*a^4*cos(c + d*x)^(1/2)*sin(c + d
*x)))/d - (8*a^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*
x)^2)^(1/2)) - (2*a^4*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(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)**(1/2),x)

[Out]

Timed out

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