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

Optimal. Leaf size=44 \[ \frac {4 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2757, 2636, 2639, 2641} \[ \frac {4 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.17, size = 39, normalized size = 0.89 \[ \frac {2 a^2 \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(2*EllipticF[(c + d*x)/2, 2] + Sin[c + d*x]/Sqrt[Cos[c + d*x]]))/d

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.50, size = 104, normalized size = 2.36 \[ -\frac {4 a^{2} \left (\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-si
n(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 \[ \int \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 82, normalized size = 1.86 \[ \frac {2\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a^2*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))

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

[Out]

Timed out

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