3.796 \(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=87 \[ \frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[Out]

6/5*a*(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+2/3*b*(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*cos(d*x+c)^(3/2)*sin(d*x+c
)/d+2/3*b*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4225, 2748, 2635, 2641, 2639} \[ \frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*b*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(3*d) + (2*a*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 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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 66, normalized size = 0.76 \[ \frac {2 \left (\sin (c+d x) \sqrt {\cos (c+d x)} (3 a \cos (c+d x)+5 b)+9 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*a*EllipticE[(c + d*x)/2, 2] + 5*b*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*b + 3*a*Cos[c + d*x]
)*Sin[c + d*x]))/(15*d)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(d*x + c)^2*sec(d*x + c) + a*cos(d*x + c)^2)*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.64, size = 262, normalized size = 3.01 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (24 a +20 b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6 a -10 b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(
24*a+20*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*a-10*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*b*(si
n(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))-9*(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))*a)/(-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 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.04, size = 80, normalized size = 0.92 \[ \frac {2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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