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

Optimal. Leaf size=189 \[ \frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}-\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[c +
d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (26*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
- (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5
*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.413719, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4264, 3823, 4022, 4013, 3808, 206} \[ \frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}-\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[c +
d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (26*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
- (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5
*d*Sqrt[a + a*Sec[c + d*x]])

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 3823

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

Rule 4022

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[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 - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 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 3808

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x)}{\sqrt{a+a \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) \sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a-4 a \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{13 a^2}{2}+a^2 \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.303028, size = 136, normalized size = 0.72 \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (2 \sqrt{1-\sec (c+d x)} \left (13 \sec ^2(c+d x)-\sec (c+d x)+3\right )+15 \sqrt{2} \sec ^{\frac{5}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{15 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(3/2)*(15*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(5/2)
+ 2*Sqrt[1 - Sec[c + d*x]]*(3 - Sec[c + d*x] + 13*Sec[c + d*x]^2))*Sin[c + d*x])/(15*d*Sqrt[1 - Sec[c + d*x]]
*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.193, size = 120, normalized size = 0.6 \begin{align*}{\frac{1}{15\,ad\sin \left ( dx+c \right ) } \left ( 15\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -6\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-28\,\cos \left ( dx+c \right ) +26 \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{\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+a*sec(d*x+c))^(1/2),x)

[Out]

1/15/d/a*(15*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-6*cos(d*x+c
)^3+8*cos(d*x+c)^2-28*cos(d*x+c)+26)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.22697, size = 482, normalized size = 2.55 \begin{align*} \text{result too large to display} \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))^(1/2),x, algorithm="maxima")

[Out]

1/60*sqrt(2)*(60*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) - 60*cos(5/2*d*x + 5/2*c)*sin(4/5*a
rctan2(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))) - 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 1) + 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(5
/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 1
) + 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))) + 60*sin(1/5*arcta
n2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/(sqrt(a)*d)

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Fricas [A]  time = 1.77105, size = 886, normalized size = 4.69 \begin{align*} \left [\frac{4 \,{\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\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 ) + \frac{15 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \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 )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{30 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{15 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\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 (a d \cos \left (d x + c\right ) + a d\right )}}\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))^(1/2),x, algorithm="fricas")

[Out]

[1/30*(4*(3*cos(d*x + c)^2 - cos(d*x + c) + 13)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin
(d*x + c) + 15*sqrt(2)*(a*cos(d*x + c) + a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sq
rt(a))/(a*d*cos(d*x + c) + a*d), 1/15*(15*sqrt(2)*(a*cos(d*x + c) + a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d
*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + 2*(3*cos(d*x + c)^2 - cos(d*x + c) +
13)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*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+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 \frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{a \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+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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