3.249 \(\int \frac{1}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=177 \[ \frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{75 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

(-75*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) - Sin[c + d*x]/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - (13*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c
+ d*x]]*(a + a*Cos[c + d*x])^(3/2)) + (49*Sin[c + d*x])/(16*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.402734, antiderivative size = 177, 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 = {2766, 2978, 2984, 12, 2782, 205} \[ \frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{75 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(-75*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) - Sin[c + d*x]/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - (13*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c
+ d*x]]*(a + a*Cos[c + d*x])^(3/2)) + (49*Sin[c + d*x])/(16*a^2*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 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 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))^{5/2}} \, dx &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{9 a}{2}-2 a \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{49 a^2}{4}-\frac{13}{2} a^2 \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int -\frac{75 a^3}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{75 \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 \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{75 \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{75 \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{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.68358, size = 506, normalized size = 2.86 \[ \frac{2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{8 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^6\left (\frac{1}{2} (c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{5}{2}\right \},\left \{1,1,\frac{11}{2}\right \},\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )}{315 \left (2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1\right )}+\frac{1}{120} \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2 \sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \csc ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (15344 \sin ^{10}\left (\frac{c}{2}+\frac{d x}{2}\right )-66122 \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right )+109737 \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-87764 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )+33980 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-5145\right )-15 \left (824 \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-2021 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )+1465 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-343\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}}\right )\right )\right )}{d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2} (a (\cos (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*((8*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2,
2, 5/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^2)/(315*(-1 + 2*
Sin[c/2 + (d*x)/2]^2)) + (Csc[c/2 + (d*x)/2]^8*(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)]*(-15*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/
2]^4*(-343 + 1465*Sin[c/2 + (d*x)/2]^2 - 2021*Sin[c/2 + (d*x)/2]^4 + 824*Sin[c/2 + (d*x)/2]^6) + Sqrt[Sin[c/2
+ (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-5145 + 33980*Sin[c/2 + (d*x)/2]^2 - 87764*Sin[c/2 + (d*x)/2]^4 +
109737*Sin[c/2 + (d*x)/2]^6 - 66122*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/2]^10)))/120))/(d*(a*(1 + Co
s[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))

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Maple [B]  time = 0.345, size = 303, normalized size = 1.7 \begin{align*}{\frac{\sqrt{2}}{32\,d{a}^{3}\sin \left ( dx+c \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}} \left ( 75\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +225\, \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 ) +225\,\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 ) +75\,\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 ) -49\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-36\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}+53\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}+32\,\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)^(5/2),x)

[Out]

1/32/d*2^(1/2)/a^3*(75*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^3*sin(d
*x+c)+225*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))+225*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))+75*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))-49*2^(1/2)*cos(d*x+c)^4-36*cos(d*x+c)^3*2^(1/2)+5
3*cos(d*x+c)^2*2^(1/2)+32*cos(d*x+c)*2^(1/2))*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/(1+cos(d*x+c))^2/cos(d*x+c)^
(3/2)

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Maxima [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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.30193, size = 563, normalized size = 3.18 \begin{align*} -\frac{75 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \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 (49 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 32\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 )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(75*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arctan(1/2*sqr
t(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)*(49*cos(d*x + c)^2 + 85*cos(d*x + c) + 32)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*c
os(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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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)**(3/2)/(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}} \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))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)