3.242 \(\int \frac{1}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.246947, 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, 2984, 12, 2782, 205} \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

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

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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{5 a}{2}-a \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int -\frac{7 a^2}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{a^3}\\ &=-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{7 \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{7 \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 )}{2 d}\\ &=-\frac{7 \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 )}{2 \sqrt{2} a^{3/2} d}-\frac{\sin (c+d x)}{2 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{5 \sin (c+d x)}{2 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.16074, size = 456, normalized size = 3.33 \[ \frac{2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{2,2,\frac{5}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )}{70 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-35}-\frac{1}{6} \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2 \sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \csc ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (124 \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-350 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )+298 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-75\right )-3 \left (34 \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-100 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )+91 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-25\right ) \tanh ^{-1}\left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}}\right )\right )\right )}{d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2} (a (\cos (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]*((4*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2,
5/2}, {1, 9/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^2)/(-35 + 70*Sin[c/2 +
(d*x)/2]^2) - (Csc[c/2 + (d*x)/2]^6*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]*(-3*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*(-25 + 91*Sin[c/2 + (d*x)/
2]^2 - 100*Sin[c/2 + (d*x)/2]^4 + 34*Sin[c/2 + (d*x)/2]^6) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)
/2]^2)]*(-75 + 298*Sin[c/2 + (d*x)/2]^2 - 350*Sin[c/2 + (d*x)/2]^4 + 124*Sin[c/2 + (d*x)/2]^6)))/6))/(d*(a*(1
+ Cos[c + d*x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.382, size = 245, normalized size = 1.8 \begin{align*}{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{4\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}} \left ( -7\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -14\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -7\,\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}- \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}-4\,\cos \left ( dx+c \right ) \sqrt{2} \right ) \sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^(3/2),x)

[Out]

1/4/d*2^(1/2)/a^2*(-7*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x
+c))-14*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-7*sin(d*x+c
)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+5*cos(d*x+c)^3*2^(1/2)-cos(d*x+c)^2*2^(
1/2)-4*cos(d*x+c)*2^(1/2))*sin(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))/(1+cos(d*x+c))^2/cos(d*x+c)^(3/
2)

________________________________________________________________________________________

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{3}{2}} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/2)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.50175, size = 471, normalized size = 3.44 \begin{align*} -\frac{7 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\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 (5 \, \cos \left (d x + c\right ) + 4\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(7*sqrt(2)*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(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)*(5*cos(d*x + c) + 4)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^
2*d*cos(d*x + c))

________________________________________________________________________________________

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)**(3/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{3}{2}} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/2)), x)