3.9.2 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx\) [802]

Optimal. Leaf size=160 \[ \frac {2 \left (7 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {20 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {20 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (7 a^2+9 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d} \]

[Out]

2/15*(7*a^2+9*b^2)*(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/
21*a*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+2/45*(7*a^2+9*b
^2)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+4/7*a*b*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*a^2*cos(d*x+c)^(7/2)*sin(d*x+c)/d+
20/21*a*b*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.15, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4349, 3873, 3854, 3856, 2720, 4130, 2719} \begin {gather*} \frac {2 \left (7 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (7 a^2+9 b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {20 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a b \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {20 a b \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2 + 9*b^2)*EllipticE[(c + d*x)/2, 2])/(15*d) + (20*a*b*EllipticF[(c + d*x)/2, 2])/(21*d) + (20*a*b*Sqr
t[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(7*a^2 + 9*b^2)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (4*a*b*Cos
[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + (2*a^2*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d)

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 3854

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 3856

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 3873

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

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4349

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 \cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^2}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {a^2+b^2 \sec ^2(c+d x)}{\sec ^{\frac {9}{2}}(c+d x)} \, dx+\left (2 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{7} \left (10 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{9} \left (\left (-7 a^2-9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {20 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (7 a^2+9 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{21} \left (10 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{15} \left (\left (-7 a^2-9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {20 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (7 a^2+9 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{21} (10 a b) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{15} \left (-7 a^2-9 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (7 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {20 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {20 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (7 a^2+9 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 113, normalized size = 0.71 \begin {gather*} \frac {84 \left (7 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+600 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} \left (7 \left (43 a^2+36 b^2\right ) \cos (c+d x)+5 a (156 b+36 b \cos (2 (c+d x))+7 a \cos (3 (c+d x)))\right ) \sin (c+d x)}{630 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(84*(7*a^2 + 9*b^2)*EllipticE[(c + d*x)/2, 2] + 600*a*b*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(7*(43*
a^2 + 36*b^2)*Cos[c + d*x] + 5*a*(156*b + 36*b*Cos[2*(c + d*x)] + 7*a*Cos[3*(c + d*x)]))*Sin[c + d*x])/(630*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(192)=384\).
time = 0.19, size = 398, normalized size = 2.49

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-1120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}+\left (2240 a^{2}+1440 b a \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-2072 a^{2}-2160 b a -504 b^{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (952 a^{2}+1680 b a +504 b^{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-168 a^{2}-480 b a -126 b^{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+150 b a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-189 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10
*a^2+(2240*a^2+1440*a*b)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2072*a^2-2160*a*b-504*b^2)*sin(1/2*d*x+1/2*
c)^6*cos(1/2*d*x+1/2*c)+(952*a^2+1680*a*b+504*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-168*a^2-480*a*b-1
26*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+150*b*a*(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))-147*(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^2-189*(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))*b^2)/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.01, size = 195, normalized size = 1.22 \begin {gather*} \frac {-150 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 150 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{3} + 90 \, a b \cos \left (d x + c\right )^{2} + 150 \, a b + 7 \, {\left (7 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 21 \, \sqrt {2} {\left (-7 i \, a^{2} - 9 i \, b^{2}\right )} {\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 ) - 21 \, \sqrt {2} {\left (7 i \, a^{2} + 9 i \, b^{2}\right )} {\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 )}{315 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(-150*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 150*I*sqrt(2)*a*b*weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(35*a^2*cos(d*x + c)^3 + 90*a*b*cos(d*x + c)^2 + 150*a
*b + 7*(7*a^2 + 9*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 21*sqrt(2)*(-7*I*a^2 - 9*I*b^2)*weierst
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*sqrt(2)*(7*I*a^2 + 9*I*b^2)*we
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [B]
time = 1.31, size = 135, normalized size = 0.84 \begin {gather*} -\frac {2\,a^2\,{\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^2\,{\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}}-\frac {4\,a\,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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*a^2*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^2*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)) - (4*a*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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