3.170 \(\int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {56 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {8 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[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*(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))/d+2/5*a^4*cos(d*x+c)^(3/2)*s
in(d*x+c)/d+2*a^4*sin(d*x+c)/d/cos(d*x+c)^(1/2)+8/3*a^4*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.12, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2757, 2636, 2639, 2641, 2635} \[ \frac {32 a^4 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {56 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {8 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(3/2),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])/(d*
Sqrt[Cos[c + d*x]]) + (8*a^4*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/
(5*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 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 {3}{2}}(c+d x)} \, dx &=\int \left (\frac {a^4}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^4}{\sqrt {\cos (c+d x)}}+6 a^4 \sqrt {\cos (c+d x)}+4 a^4 \cos ^{\frac {3}{2}}(c+d x)+a^4 \cos ^{\frac {5}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+a^4 \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (4 a^4\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {12 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)}{d \sqrt {\cos (c+d x)}}+\frac {8 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \left (3 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx-a^4 \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^4\right ) \int \frac {1}{\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)}{d \sqrt {\cos (c+d x)}}+\frac {8 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 6.20, size = 497, normalized size = 4.18 \[ -\frac {7 \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )}{\sqrt {\tan ^2(c)+1} \sqrt {1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt {\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac {\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt {\tan ^2(c)+1}}+\frac {2 \cos ^2(c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{20 d}-\frac {2 \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin (c) \left (-\sqrt {\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{3 d \sqrt {\cot ^2(c)+1}}+\sqrt {\cos (c+d x)} \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac {\sin (c) \cos (d x)}{6 d}+\frac {\sin (2 c) \cos (2 d x)}{80 d}+\frac {\cos (c) \sin (d x)}{6 d}+\frac {\cos (2 c) \sin (2 d x)}{80 d}+\frac {\sec (c) \sin (d x) \sec (c+d x)}{8 d}-\frac {(33 \cos (2 c)+23) \csc (c) \sec (c)}{80 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(3/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(-1/80*((23 + 33*Cos[2*c])*Csc[c]*Sec[c])/d + (
Cos[d*x]*Sin[c])/(6*d) + (Cos[2*d*x]*Sin[2*c])/(80*d) + (Cos[c]*Sin[d*x])/(6*d) + (Sec[c]*Sec[c + d*x]*Sin[d*x
])/(8*d) + (Cos[2*c]*Sin[2*d*x])/(80*d)) - (2*(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]]]])/(3*d*Sq
rt[1 + Cot[c]^2]) - (7*(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[Tan[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]]))/(20*d)

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(3/2),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)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(3/2), x)

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maple [A]  time = 0.52, size = 194, normalized size = 1.63 \[ -\frac {8 a^{4} \left (-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+26 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-19 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4/cos(d*x+c)^(3/2),x)

[Out]

-8/15*a^4*(-6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+26*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+20*(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))-21*(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))-19*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c))/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 \[ \int \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(3/2), x)

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mupad [B]  time = 0.79, size = 149, normalized size = 1.25 \[ \frac {12\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {32\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {8\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^4/cos(c + d*x)^(3/2),x)

[Out]

(12*a^4*ellipticE(c/2 + (d*x)/2, 2))/d + (32*a^4*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (8*a^4*cos(c + d*x)^(1/2
)*sin(c + d*x))/(3*d) + (2*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)
*(sin(c + d*x)^2)^(1/2)) - (2*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))

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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)**(3/2),x)

[Out]

Timed out

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