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

Optimal. Leaf size=152 \[ -\frac{2 b \left (a^2+3 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^4 d}+\frac{2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac{2 b^4 \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d} \]

[Out]

(2*(3*a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*b*(a^2 + 3*b^2)*EllipticF[(c + d*x)/2, 2])/(3*a^4
*d) + (2*b^4*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^4*(a + b)*d) - (2*b*Sqrt[Cos[c + d*x]]*Sin[c + d*x]
)/(3*a^2*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d)

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Rubi [A]  time = 0.602274, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4264, 3853, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{2 b \left (a^2+3 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac{2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac{2 b^4 \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*b*(a^2 + 3*b^2)*EllipticF[(c + d*x)/2, 2])/(3*a^4
*d) + (2*b^4*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^4*(a + b)*d) - (2*b*Sqrt[Cos[c + d*x]]*Sin[c + d*x]
)/(3*a^2*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*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 3853

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

Rule 4104

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

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 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 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+b \sec (c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\\ &=\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{5 b}{2}+\frac{3}{2} a \sec (c+d x)+\frac{3}{2} b \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{5 a}\\ &=-\frac{2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{4} \left (3 a^2+5 b^2\right )-a b \sec (c+d x)+\frac{5}{4} b^2 \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 a^2}\\ &=-\frac{2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{4} a \left (3 a^2+5 b^2\right )-\left (a^2 b-\frac{3}{4} b \left (3 a^2+5 b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^4}+\frac{\left (b^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{b^4 \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^4}-\frac{\left (b \left (a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{3 a^4}+\frac{\left (\left (3 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 a^3}\\ &=\frac{2 b^4 \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}-\frac{2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac{\left (b \left (a^2+3 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^4}+\frac{\left (3 a^2+5 b^2\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3}\\ &=\frac{2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 b \left (a^2+3 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac{2 b^4 \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}-\frac{2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 1.58215, size = 228, normalized size = 1.5 \[ \frac{-\frac{6 \left (3 a^2+5 b^2\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+8 b \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )+\frac{2 \left (9 a^2+5 b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+4 \sin (c+d x) \sqrt{\cos (c+d x)} (3 a \cos (c+d x)-5 b)}{30 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(9*a^2 + 5*b^2)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 8*b*(2*EllipticF[(c + d*x)/2, 2] - (2
*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*Sqrt[Cos[c + d*x]]*(-5*b + 3*a*Cos[c + d*x])*Sin[c
+ d*x] - (6*(3*a^2 + 5*b^2)*(2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] - 2*b*(a + b)*EllipticF[ArcSin[Sq
rt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2
*b*Sqrt[Sin[c + d*x]^2]))/(30*a^2*d)

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Maple [B]  time = 1.995, size = 668, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*a^4+24*a^3*b)*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^6+(24*a^4-44*a^3*b+20*a^2*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*a^4+16*a^3*b-10*a^2*b^2)*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+5*a^2*b^2*(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))-15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3+15*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(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(co
s(1/2*d*x+1/2*c),2^(1/2))*a^4+9*(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^3*b-15*(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*b^2+15*(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*b^3-15*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(co
s(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/a^4/(a-b)/(-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}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a), 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}}}{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a), 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+b*sec(d*x+c)),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}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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