3.794 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=135 \[ \frac {14 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {14 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {10 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

[Out]

14/15*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+10/21*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+14/45*a*cos(d*x+c)^(3/2)*sin
(d*x+c)/d+2/7*b*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*a*cos(d*x+c)^(7/2)*sin(d*x+c)/d+10/21*b*sin(d*x+c)*cos(d*x+c
)^(1/2)/d

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Rubi [A]  time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {14 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {14 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {10 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x]),x]

[Out]

(14*a*EllipticE[(c + d*x)/2, 2])/(15*d) + (10*b*EllipticF[(c + d*x)/2, 2])/(21*d) + (10*b*Sqrt[Cos[c + d*x]]*S
in[c + d*x])/(21*d) + (14*a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (2*b*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7
*d) + (2*a*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*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 {9}{2}}(c+d x) (a+b \sec (c+d x)) \, dx &=\int \cos ^{\frac {7}{2}}(c+d x) (b+a \cos (c+d x)) \, dx\\ &=a \int \cos ^{\frac {9}{2}}(c+d x) \, dx+b \int \cos ^{\frac {7}{2}}(c+d x) \, dx\\ &=\frac {2 b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{9} (7 a) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{7} (5 b) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {10 b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {14 a \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{15} (7 a) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} (5 b) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {14 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {10 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {10 b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {14 a \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 90, normalized size = 0.67 \[ \frac {\sqrt {\cos (c+d x)} (266 a \sin (2 (c+d x))+35 a \sin (4 (c+d x))+690 b \sin (c+d x)+90 b \sin (3 (c+d x)))+1176 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+600 b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x]),x]

[Out]

(1176*a*EllipticE[(c + d*x)/2, 2] + 600*b*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(690*b*Sin[c + d*x] +
266*a*Sin[2*(c + d*x)] + 90*b*Sin[3*(c + d*x)] + 35*a*Sin[4*(c + d*x)]))/(1260*d)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^4*sec(d*x + c) + a*cos(d*x + c)^4)*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)*cos(d*x + c)^(9/2), x)

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maple [A]  time = 4.02, size = 318, normalized size = 2.36 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-1120 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2240 a +720 b \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-2072 a -1080 b \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (952 a +840 b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-168 a -240 b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+75 b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c)),x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
10+(2240*a+720*b)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2072*a-1080*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/
2*c)+(952*a+840*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-168*a-240*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+75*b*(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))-1
47*(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 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)*cos(d*x + c)^(9/2), x)

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mupad [B]  time = 1.31, size = 87, normalized size = 0.64 \[ -\frac {2\,a\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(9/2)*(a + b/cos(c + d*x)),x)

[Out]

- (2*a*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(
1/2)) - (2*b*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*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)**(9/2)*(a+b*sec(d*x+c)),x)

[Out]

Timed out

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