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

Optimal. Leaf size=100 \[ \frac {5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \sec (c+d x)+a)} \]

[Out]

-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))/a/d+5/3*(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))/a/d+5/3*sin(d*x+c)*cos(d*x+c)^(1/2)/a
/d-sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4264, 3819, 3787, 3769, 3771, 2641, 2639} \[ \frac {5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \sec (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (5*Sqrt[Cos[c + d*x]]*Sin[c + d
*x])/(3*a*d) - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [C]  time = 2.15, size = 292, normalized size = 2.92 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {4 \sin (c) \cos (d x)+4 \cos (c) \sin (d x)+6 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+12 \cot (c)+6 \csc (c)}{d \sqrt {\cos (c+d x)}}-\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+9 \left (1+e^{2 i (c+d x)}\right )\right ) \sec (c+d x)}{\left (-1+e^{2 i c}\right ) d \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*(((-2*I)*Sqrt[2]*(9*(1 + E^((2*I)*(c + d*x))) + 9*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1
+ E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x])/(d*E^(I*(c + d*x
))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (12*Cot[c] + 6*Csc[c] + 4*Cos[d*x]*Si
n[c] + 6*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] + 4*Cos[c]*Sin[d*x])/(d*Sqrt[Cos[c + d*x]])))/(3*a*(1 + Sec[c
+ d*x]))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.93, size = 215, normalized size = 2.15 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2)))-8*sin(1/2*d*x+1/2*c)^6+18*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/a/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 \[ \int \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{\frac {3}{2}}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)**(3/2)/(sec(c + d*x) + 1), x)/a

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