3.248 \(\int \frac{1}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{9 \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.260541, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2766, 2978, 12, 2782, 205} \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{9 \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[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]])])/(16*Sqrt[2]*a^(5/2)*
d) - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (9*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/
(16*a*d*(a + a*Cos[c + d*x])^(3/2))

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 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 12

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

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 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]

Rubi steps

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

Mathematica [A]  time = 1.26198, size = 134, normalized size = 0.98 \[ -\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\cos (c+d x) (9 \cos (c+d x)+13) \sqrt{2-2 \sec (c+d x)}-76 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{\sin ^2\left (\frac{1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )\right )}{32 \sqrt{2} a^2 d \sqrt{\cos (c+d x)-1} \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

-(Sec[(c + d*x)/2]^2*(-76*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^4 + Cos[c + d*x]*
(13 + 9*Cos[c + d*x])*Sqrt[2 - 2*Sec[c + d*x]])*Tan[(c + d*x)/2])/(32*Sqrt[2]*a^2*d*Sqrt[-1 + Cos[c + d*x]]*Sq
rt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 0.332, size = 245, normalized size = 1.8 \begin{align*}{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }{32\,d{a}^{3} \left ( 1+\cos \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 19\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +38\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}+19\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) -4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}+13\,\cos \left ( dx+c \right ) \sqrt{2} \right ){\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d*2^(1/2)/a^3*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))*(19*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin((-1+c
os(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+38*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin((-1+cos(d*x+c))/sin
(d*x+c))*sin(d*x+c)*cos(d*x+c)-9*cos(d*x+c)^3*2^(1/2)+19*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*sin(d*x+c)-4*cos(d*x+c)^2*2^(1/2)+13*cos(d*x+c)*2^(1/2))/(1+cos(d*x+c))/cos(d*x+c)^(1/2)/sin(d
*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.29245, size = 495, normalized size = 3.61 \begin{align*} \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{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (9 \, \cos \left (d x + c\right ) + 13\right )} \sqrt{\cos \left (d x + c\right )} \sin \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(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(1/2*sqrt(2)*sqrt(a*co
s(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*sqrt(a*cos(d*
x + c) + a)*(9*cos(d*x + c) + 13)*sqrt(cos(d*x + c))*sin(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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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(1/cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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