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

Optimal. Leaf size=147 \[ \frac {20 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {32 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {4 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {32 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {20 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

[Out]

32/15*a^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^2*(c
os(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+32/45*a^2*cos(d*x+c)^(3/
2)*sin(d*x+c)/d+4/7*a^2*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^2*sin(d*x+
c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.17, antiderivative size = 147, 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 = {4264, 3788, 3769, 3771, 2641, 4045, 2639} \[ \frac {20 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {32 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {4 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {32 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {20 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a^2*EllipticE[(c + d*x)/2, 2])/(15*d) + (20*a^2*EllipticF[(c + d*x)/2, 2])/(21*d) + (20*a^2*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(21*d) + (32*a^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (4*a^2*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 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 3788

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 4045

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 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 \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \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+a^2 \sec ^2(c+d x)}{\sec ^{\frac {9}{2}}(c+d x)} \, dx+\left (2 a^2 \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^2 \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^2 \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 (16 a^2 \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^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {32 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a^2 \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^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (16 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {20 a^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {32 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a^2 \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^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{15} \left (16 a^2\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {32 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {20 a^2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {20 a^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {32 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a^2 \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 [C]  time = 6.12, size = 548, normalized size = 3.73 \[ -\frac {4 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )}{\sqrt {\tan ^2(c)+1} \sqrt {1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt {\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac {\frac {\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt {\tan ^2(c)+1}}+\frac {2 \cos ^2(c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt {\cos (c) \sqrt {\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{15 d}-\frac {5 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \sec (c+d x)+a)^2 \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin (c) \left (-\sqrt {\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{21 d \sqrt {\cot ^2(c)+1}}+\cos ^{\frac {5}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac {23 \sin (c) \cos (d x)}{84 d}+\frac {37 \sin (2 c) \cos (2 d x)}{360 d}+\frac {\sin (3 c) \cos (3 d x)}{28 d}+\frac {\sin (4 c) \cos (4 d x)}{144 d}+\frac {23 \cos (c) \sin (d x)}{84 d}+\frac {37 \cos (2 c) \sin (2 d x)}{360 d}+\frac {\cos (3 c) \sin (3 d x)}{28 d}+\frac {\cos (4 c) \sin (4 d x)}{144 d}-\frac {8 \cot (c)}{15 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Cos[c + d*x]^(5/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*((-8*Cot[c])/(15*d) + (23*Cos[d*x]*Sin[c])/(84*
d) + (37*Cos[2*d*x]*Sin[2*c])/(360*d) + (Cos[3*d*x]*Sin[3*c])/(28*d) + (Cos[4*d*x]*Sin[4*c])/(144*d) + (23*Cos
[c]*Sin[d*x])/(84*d) + (37*Cos[2*c]*Sin[2*d*x])/(360*d) + (Cos[3*c]*Sin[3*d*x])/(28*d) + (Cos[4*c]*Sin[4*d*x])
/(144*d)) - (5*Cos[c + d*x]^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
+ (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt
[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*Sqrt[1 + Cot[c]^2
]) - (4*Cos[c + d*x]^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*((HypergeometricPFQ[{-1/2, -1/4}, {3
/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[
1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) -
((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c
]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(15*d)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*cos(d*x + c)^4*sec(d*x + c)^2 + 2*a^2*cos(d*x + c)^4*sec(d*x + c) + a^2*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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.62, size = 260, normalized size = 1.77 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (560 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-960 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+608 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-205 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+93 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(560*cos(1/2*d*x+1/2*c)^11-960*cos(1/2*d*x+
1/2*c)^9+608*cos(1/2*d*x+1/2*c)^7-96*cos(1/2*d*x+1/2*c)^5-205*cos(1/2*d*x+1/2*c)^3+75*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+93*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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.18, size = 136, normalized size = 0.93 \[ -\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*a^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^2*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)) - (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))

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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+a*sec(d*x+c))**2,x)

[Out]

Timed out

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