3.1247 \(\int \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=129 \[ \frac{2 a (7 A+5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 (A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{5 d} \]

[Out]

(2*a*(7*A + 5*B + 15*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*(A + 5*B)*Sqrt[C
os[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.468495, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {4265, 4086, 4013, 3804} \[ \frac{2 a (7 A+5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 (A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*A + 5*B + 15*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*(A + 5*B)*Sqrt[C
os[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(5*d)

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4086

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*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 4013

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

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rubi steps

\begin{align*} \int \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (A+5 B)+\frac{1}{2} a (2 A+5 C) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 (A+5 B) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{1}{15} \left ((7 A+5 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a (7 A+5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 (A+5 B) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.188029, size = 82, normalized size = 0.64 \[ \frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} \left ((4 A+5 B) \cos (c+d x)+3 A \cos ^2(c+d x)+8 A+10 B+15 C\right )}{15 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*(8*A + 10*B + 15*C + (4*A + 5*B)*Cos[c + d*x] + 3*A*Cos[c + d*x]^2)*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(15*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.351, size = 89, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 3\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,A\cos \left ( dx+c \right ) +5\,B\cos \left ( dx+c \right ) +8\,A+10\,B+15\,C \right ) }{15\,d\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \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)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+8*A+10*B+15*C)*cos(d*x+c)^(1/2)*(a*(co
s(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.23529, size = 433, normalized size = 3.36 \begin{align*} \frac{\sqrt{2}{\left (30 \, \cos \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \cos \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 30 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )} A \sqrt{a} + 120 \, \sqrt{2} C \sqrt{a} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) - 10 \,{\left (3 \, \sqrt{2} \cos \left (\frac{3}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) \sin \left (d x + c\right ) -{\left (3 \, \sqrt{2} \cos \left (d x + c\right ) + 2 \, \sqrt{2}\right )} \sin \left (\frac{3}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) - 3 \, \sqrt{2} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )\right )} B \sqrt{a}}{60 \, d} \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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/60*(sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/
5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/
2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 120*sqrt(2)*C*sqrt(a)*
sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 10*(3*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(
d*x + c) - (3*sqrt(2)*cos(d*x + c) + 2*sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*sqrt(2)*sin(1
/2*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a))/d

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Fricas [A]  time = 0.480971, size = 225, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (3 \, A \cos \left (d x + c\right )^{2} +{\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 10 \, B + 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right ) + d\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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*A*cos(d*x + c)^2 + (4*A + 5*B)*cos(d*x + c) + 8*A + 10*B + 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)

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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)*(a+a*sec(d*x+c))**(1/2),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 )} \sqrt{a \sec \left (d x + c\right ) + a} \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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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