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

Optimal. Leaf size=160 \[ -\frac{5 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a^2 d}+\frac{56 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}+\frac{56 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 \sin (c+d x) \sqrt{\cos (c+d x)}}{a^2 d}-\frac{3 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

(56*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*EllipticF[(c + d*x)/2, 2])/(a^2*d) - (5*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(a^2*d) + (56*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*a^2*d) - (3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(a^2*
d*(1 + Sec[c + d*x])) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.283813, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{56 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}+\frac{56 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 \sin (c+d x) \sqrt{\cos (c+d x)}}{a^2 d}-\frac{3 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(56*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*EllipticF[(c + d*x)/2, 2])/(a^2*d) - (5*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(a^2*d) + (56*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*a^2*d) - (3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(a^2*
d*(1 + Sec[c + d*x])) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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))^2} \, 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))^2} \, dx\\ &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{11 a}{2}+\frac{7}{2} a \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-28 a^2+\frac{45}{2} a^2 \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left (15 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}+\frac{\left (28 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^2}\\ &=-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d}+\frac{56 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left (5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{2 a^2}+\frac{\left (28 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 a^2}\\ &=-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d}+\frac{56 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{5 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a^2}+\frac{28 \int \sqrt{\cos (c+d x)} \, dx}{5 a^2}\\ &=\frac{56 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d}+\frac{56 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 2.75431, size = 366, normalized size = 2.29 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{2 \left (-40 \sin (c) \cos (d x)+6 \sin (2 c) \cos (2 d x)-40 \cos (c) \sin (d x)+6 \cos (2 c) \sin (2 d x)+5 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right )+5 \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )-120 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right )-216 \cot (c)-120 \csc (c)\right )}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 i \sqrt{2} e^{-i (c+d x)} \sec ^2(c+d x) \left (56 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+25 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+56 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{5 a^2 (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*(((4*I)*Sqrt[2]*(56*(1 + E^((2*I)*(c + d*x))) + 56*(-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))] + 25*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt
[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^2)/(d*E^(I*(c +
d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (2*(-216*Cot[c] - 120*Csc[c] - 40
*Cos[d*x]*Sin[c] + 6*Cos[2*d*x]*Sin[2*c] - 120*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] + 5*Sec[c/2]*Sec[(c + d*
x)/2]^3*Sin[(d*x)/2] - 40*Cos[c]*Sin[d*x] + 6*Cos[2*c]*Sin[2*d*x] + 5*Sec[(c + d*x)/2]^2*Tan[c/2]))/(3*d*Cos[c
+ d*x]^(3/2))))/(5*a^2*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 1.534, size = 283, normalized size = 1.8 \begin{align*} -{\frac{1}{30\,{a}^{2}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 ( 96\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}-352\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+120\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-150\,\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 ) ^{3}-336\,\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 ) ^{3}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +266\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-135\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+5 \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\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)^(5/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*cos(1/2*d*x+1/2*c)^10-352*cos(1/2*d*x+1/2*c)
^8+120*cos(1/2*d*x+1/2*c)^6-150*(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)^3-336*(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)^3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+266*cos(1/2*d*x+1/2*c)^4-135*cos(1/2*d*x+1/2*c)^2+
5)/a^2/cos(1/2*d*x+1/2*c)^3/(-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., 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 )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

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

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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^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^(5/2)/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), 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))**2,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 )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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