3.1181 \(\int \cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=93 \[ \frac{2 B \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.116145, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4064, 3023, 2748, 2639, 2635, 2641} \[ \frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4064

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sqrt{\cos (c+d x)} \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2}{5} \int \sqrt{\cos (c+d x)} \left (\frac{1}{2} (3 A+5 C)+\frac{5}{2} B \cos (c+d x)\right ) \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+B \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{5} (3 A+5 C) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{1}{3} B \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.287705, size = 72, normalized size = 0.77 \[ \frac{2 \left (5 B \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} (3 A \cos (c+d x)+5 B)+3 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2] + 5*B*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*B + 3*A*Co
s[c + d*x])*Sin[c + d*x]))/(15*d)

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Maple [B]  time = 2.305, size = 308, normalized size = 3.3 \begin{align*} -{\frac{2}{15\,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 ( -24\,A \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) + \left ( 24\,A+20\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -6\,A-10\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -9\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +5\,B\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -15\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(
24*A+20*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*A-10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-9*A*(si
n(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))+5*B*(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*C*(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)))/(-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{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \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)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c) + A*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)+C*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 (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \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)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(5/2), x)