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

Optimal. Leaf size=136 \[ \frac {3 a \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3 \sqrt {a} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d} \]

[Out]

3/4*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/2*a*sin(d
*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+3/4*a*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4264, 3803, 3801, 215} \[ \frac {3 a \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3 \sqrt {a} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(4*
d) + (a*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a*Sin[c + d*x])/(4*d*Cos[c + d*x]
^(3/2)*Sqrt[a + a*Sec[c + d*x]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 3803

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

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1}{4} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {3 a \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1}{8} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {3 a \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac {3 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {3 a \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 100, normalized size = 0.74 \[ \frac {2 a \sin (c+d x) \left (\frac {1}{8} \cos (c+d x) (3 \cos (c+d x)+2)+\frac {3 \sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )}{8 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}\right )}{d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*((Cos[c + d*x]*(2 + 3*Cos[c + d*x]))/8 + (3*ArcSin[Sqrt[1 - Sec[c + d*x]]])/(8*Sqrt[1 - Sec[c + d*x]]*Sec
[c + d*x]^(5/2)))*Sin[c + d*x])/(d*Cos[c + d*x]^(7/2)*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.55, size = 355, normalized size = 2.61 \[ \left [\frac {4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/16*(4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(3*cos(d*x + c) + 2)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*(cos
(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
)*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c
)^2)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 1/8*(2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(3*cos(d*x + c) +
2)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*
cos(d*x + c)^3 + d*cos(d*x + c)^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sec \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.01, size = 213, normalized size = 1.57 \[ \frac {\left (3 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-3 \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+6 \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+4 \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{16 d \cos \left (d x +c \right )^{\frac {3}{2}} \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x)

[Out]

1/16/d*(3*cos(d*x+c)^2*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))-3*cos(d
*x+c)^2*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))+6*(-2/(1+cos(d*x+c)))^
(1/2)*cos(d*x+c)*sin(d*x+c)+4*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2/(1
+cos(d*x+c)))^(1/2)/cos(d*x+c)^(3/2)/sin(d*x+c)^2*(cos(d*x+c)^2-1)

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maxima [B]  time = 0.72, size = 1264, normalized size = 9.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-1/16*(12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(d*x + c), cos(d*x + c)))
+ 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 4*
(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c))) - 12*(sqr
t(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*(2*(2*cos
(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4
*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x +
c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x +
c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x + 2*c) +
1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(
2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)
))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)
)) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4
*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4
*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2
*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*s
in(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x
+ 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c
)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x
+ c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(s
in(d*x + c), cos(d*x + c))) + 2) - 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(7/
2*arctan2(sin(d*x + c), cos(d*x + c))) - 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*s
in(5/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(
2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) +
sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*sqrt(a)/((2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c)
+ cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin
(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(1/2)/cos(c + d*x)^(5/2),x)

[Out]

int((a + a/cos(c + d*x))^(1/2)/cos(c + d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(1/2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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