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

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

[Out]

(32*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)

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Rubi [A]  time = 0.174267, antiderivative size = 147, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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*}

Mathematica [C]  time = 6.15495, size = 548, normalized size = 3.73 \[ -\frac{4 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{15 d}-\frac{5 \csc (c) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{21 d \sqrt{\cot ^2(c)+1}}+\cos ^{\frac{5}{2}}(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac{23 \sin (c) \cos (d x)}{84 d}+\frac{37 \sin (2 c) \cos (2 d x)}{360 d}+\frac{\sin (3 c) \cos (3 d x)}{28 d}+\frac{\sin (4 c) \cos (4 d x)}{144 d}+\frac{23 \cos (c) \sin (d x)}{84 d}+\frac{37 \cos (2 c) \sin (2 d x)}{360 d}+\frac{\cos (3 c) \sin (3 d x)}{28 d}+\frac{\cos (4 c) \sin (4 d x)}{144 d}-\frac{8 \cot (c)}{15 d}\right ) \]

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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Maple [A]  time = 1.381, size = 260, normalized size = 1.8 \begin{align*} -{\frac{4\,{a}^{2}}{315\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 560\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}-960\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+608\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-96\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-205\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+75\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -168\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +93\,\cos \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}

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)