3.389 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=207 \[ -\frac{21 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{2 a^3 d}+\frac{231 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{77 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a^3 d}-\frac{21 \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a^3 d}-\frac{63 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

(231*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - (21*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) - (21*Sqrt[Cos[c + d*x]]
*Sin[c + d*x])/(2*a^3*d) + (77*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(10*a^3*d) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])
/(5*d*(a + a*Sec[c + d*x])^3) - (4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^2) - (63*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.428436, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4264, 3817, 4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac{21 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{231 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{77 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a^3 d}-\frac{21 \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a^3 d}-\frac{63 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)/(a + a*Sec[c + d*x])^3,x]

[Out]

(231*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - (21*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) - (21*Sqrt[Cos[c + d*x]]
*Sin[c + d*x])/(2*a^3*d) + (77*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(10*a^3*d) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])
/(5*d*(a + a*Sec[c + d*x])^3) - (4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^2) - (63*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

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

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\\ &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{15 a}{2}+\frac{9}{2} a \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{105 a^2}{2}+42 a^2 \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{63 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1155 a^3}{4}+\frac{945}{4} a^3 \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{63 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (63 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{4 a^3}+\frac{\left (77 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{4 a^3}\\ &=-\frac{21 \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a^3 d}+\frac{77 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a^3 d}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{63 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{4 a^3}+\frac{\left (231 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{21 \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a^3 d}+\frac{77 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a^3 d}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{63 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{21 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}+\frac{231 \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{231 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{21 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac{21 \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a^3 d}+\frac{77 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a^3 d}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{63 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.03894, size = 391, normalized size = 1.89 \[ \frac{2 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\frac{1}{16} \sec \left (\frac{c}{2}\right ) \left (770 \sin \left (c+\frac{d x}{2}\right )-840 \sin \left (c+\frac{3 d x}{2}\right )+150 \sin \left (2 c+\frac{3 d x}{2}\right )-238 \sin \left (2 c+\frac{5 d x}{2}\right )-40 \sin \left (3 c+\frac{5 d x}{2}\right )-5 \sin \left (3 c+\frac{7 d x}{2}\right )-5 \sin \left (4 c+\frac{7 d x}{2}\right )+\sin \left (4 c+\frac{9 d x}{2}\right )+\sin \left (5 c+\frac{9 d x}{2}\right )-1210 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )-264 \cot (c)-198 \csc (c)}{\cos ^{\frac{5}{2}}(c+d x)}+\frac{42 i \sqrt{2} e^{-i (c+d x)} \sec ^3(c+d x) \left (11 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+11 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{5 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(5/2)/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]^6*(((42*I)*Sqrt[2]*(11*(1 + E^((2*I)*(c + d*x))) + 11*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)
*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sq
rt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^3)/(E^(I*(c +
d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (-264*Cot[c] - 198*Csc[c] + (Sec[
c/2]*Sec[(c + d*x)/2]^5*(-1210*Sin[(d*x)/2] + 770*Sin[c + (d*x)/2] - 840*Sin[c + (3*d*x)/2] + 150*Sin[2*c + (3
*d*x)/2] - 238*Sin[2*c + (5*d*x)/2] - 40*Sin[3*c + (5*d*x)/2] - 5*Sin[3*c + (7*d*x)/2] - 5*Sin[4*c + (7*d*x)/2
] + Sin[4*c + (9*d*x)/2] + Sin[5*c + (9*d*x)/2]))/16)/Cos[c + d*x]^(5/2)))/(5*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 1.726, size = 296, normalized size = 1.4 \begin{align*} -{\frac{1}{20\,{a}^{3}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 ( 64\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}-288\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}-76\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-210\,\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 ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-462\,\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} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +530\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-248\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+19\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \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 ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5} \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)^(5/2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(64*cos(1/2*d*x+1/2*c)^12-288*cos(1/2*d*x+1/2*c)
^10-76*cos(1/2*d*x+1/2*c)^8-210*(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))*cos(1/2*d*x+1/2*c)^5-462*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)
*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+530*cos(1/2*d*x+1/2*c)^6-248*cos(1/2*d*x+1/2*c)^4+
19*cos(1/2*d*x+1/2*c)^2-1)/a^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5/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)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^(5/2)/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + a^3), 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)**(5/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)/(a*sec(d*x + c) + a)^3, x)