3.2.64 \(\int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [164]

Optimal. Leaf size=91 \[ -\frac {4 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {20 a^3 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

-4*a^3*(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+20/3*a^3*(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+2/3*a^3*sin(d*x+c)/d/cos(d*x
+c)^(3/2)+6*a^3*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2836, 2716, 2720, 2719} \begin {gather*} \frac {20 a^3 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(5/2),x]

[Out]

(-4*a^3*EllipticE[(c + d*x)/2, 2])/d + (20*a^3*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^3*Sin[c + d*x])/(3*d*Co
s[c + d*x]^(3/2)) + (6*a^3*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])

Rule 2716

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 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 2836

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))^3}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\int \left (\frac {a^3}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {3 a^3}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {3 a^3}{\sqrt {\cos (c+d x)}}+a^3 \sqrt {\cos (c+d x)}\right ) \, dx\\ &=a^3 \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx+a^3 \int \sqrt {\cos (c+d x)} \, dx+\left (3 a^3\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\left (3 a^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {6 a^3 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{3} a^3 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {20 a^3 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.26, size = 463, normalized size = 5.09 \begin {gather*} \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(-5+\cos (2 c)) \csc (c) \sec (c)}{8 d}+\frac {\sec (c) \sec ^2(c+d x) \sin (d x)}{12 d}+\frac {\sec (c) \sec (c+d x) (\sin (c)+9 \sin (d x))}{12 d}\right )-\frac {5 (a+a \cos (c+d x))^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{6 d \sqrt {1+\cot ^2(c)}}+\frac {(a+a \cos (c+d x))^3 \csc (c) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{4 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(5/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/8*((-5 + Cos[2*c])*Csc[c]*Sec[c])/d + (Sec[
c]*Sec[c + d*x]^2*Sin[d*x])/(12*d) + (Sec[c]*Sec[c + d*x]*(Sin[c] + 9*Sin[d*x]))/(12*d)) - (5*(a + a*Cos[c + d
*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(6*d*Sqrt[1 + Cot[c]^2]) + ((a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (
d*x)/2]^6*((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]]))/(4*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(370\) vs. \(2(135)=270\).
time = 0.21, size = 371, normalized size = 4.08

method result size
default \(-\frac {4 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (18 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{3 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(371\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c
)^2+1)/sin(1/2*d*x+1/2*c)^3*(18*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-10*(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))*sin(1/2*d*x+1/2*c)^2-6*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-10*sin(1/2*
d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(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))+3*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 187, normalized size = 2.05 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\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 ) - 3 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\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 ) - {\left (9 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(5*I*sqrt(2)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*a
^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*a^3*cos(d*x + c)^2*w
eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*a^3*cos(d*x + c
)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (9*a^3*cos(d*x + c) +
a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3/cos(d*x+c)**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5990 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)

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Mupad [B]
time = 1.02, size = 126, normalized size = 1.38 \begin {gather*} \frac {2\,\left (a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {6\,a^3\,\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^3\,\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^3/cos(c + d*x)^(5/2),x)

[Out]

(2*(a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*a^3*ellipticF(c/2 + (d*x)/2, 2)))/d + (6*a^3*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^3*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))

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