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

Optimal. Leaf size=116 \[ \frac{2 b \left (a^2+b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}+\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 b \sin (c+d x) \sqrt{\cos (c+d x)}}{5 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

[Out]

(6*a*(a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a^2*b
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^2*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.227948, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4264, 3841, 4047, 3771, 2639, 4045, 2641} \[ \frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 b \sin (c+d x) \sqrt{\cos (c+d x)}}{5 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*(a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a^2*b
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^2*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*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 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 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 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 \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \sec (c+d x))^3}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{6 a^2 b+\frac{3}{2} a \left (a^2+5 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (a^2+5 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{6 a^2 b+\frac{1}{2} b \left (a^2+5 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{5} \left (3 a \left (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\\ &=\frac{8 a^2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \left (3 a \left (a^2+5 b^2\right )\right ) \int \sqrt{\cos (c+d x)} \, dx+\left (b \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\left (b \left (a^2+b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{8 a^2 b \sqrt{\cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.364365, size = 84, normalized size = 0.72 \[ \frac{2 \left (5 b \left (a^2+b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 \left (a^3+5 a b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+a^2 \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+5 b)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.634, size = 376, normalized size = 3.2 \begin{align*} -{\frac{2}{5\,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 ( -8\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 8\,{a}^{3}+20\,{a}^{2}b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -2\,{a}^{3}-10\,{a}^{2}b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +5\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}b+5\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{3}-3\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}-15\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a{b}^{2} \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)^(5/2)*(a+b*sec(d*x+c))^3,x)

[Out]

-2/5*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(
8*a^3+20*a^2*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2*a^3-10*a^2*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*
b+5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^3-3*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-15*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2)/(-2*si
n(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)^(5/2)*(a+b*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 ({\left (b^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{2}\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)^(5/2)*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((b^3*cos(d*x + c)^2*sec(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2*sec(d*x + c)^2 + 3*a^2*b*cos(d*x + c)^2*s
ec(d*x + c) + a^3*cos(d*x + c)^2)*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)**(5/2)*(a+b*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{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{5}{2}}\,{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))^3,x, algorithm="giac")

[Out]

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