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

Optimal. Leaf size=157 \[ \frac {19 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {9 \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2)-9/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c
)^(1/2)+19/32*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*
sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4222, 2766, 2978, 12, 2782, 205} \[ \frac {19 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {9 \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(19*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c + d*x]]*Sq
rt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - Sin[c + d*x]/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) -
(9*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 4222

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

Rubi steps

\begin {align*} \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {7 a}{2}-a \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {9 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {19 a^2}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {9 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (19 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {9 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (19 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac {19 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {9 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.94, size = 131, normalized size = 0.83 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (76 \tanh ^{-1}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )-\cos (c+d x) (9 \cos (c+d x)+13) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {2-2 \sec (c+d x)}\right )}{64 \sqrt {2} a^2 d \sqrt {1-\sec (c+d x)} \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((76*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]] - Cos[c + d*x]*(13 + 9*Cos[c + d*x])*Sec[(c + d*x)/2]^4
*Sqrt[2 - 2*Sec[c + d*x]])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(64*Sqrt[2]*a^2*d*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[
1 - Sec[c + d*x]])

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fricas [A]  time = 1.10, size = 169, normalized size = 1.08 \[ -\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (9 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d
*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*(9*cos(d*x + c)^2 + 13*co
s(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x
+ c) + a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 222, normalized size = 1.41 \[ \frac {\sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (9 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-19 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+4 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-19 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-13 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {2}}{32 d \sin \left (d x +c \right )^{5} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^2*(9*cos(d*x+c)^2*2^(1/2)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)-19*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+4*cos(d*x+c)*2^(1/2)
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-19*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-13*2^(1/2)*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2))/sin(d*x+c)^5/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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