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

Optimal. Leaf size=157 \[ -\frac {7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {5 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-1/2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(3/2)-7/4*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^(3/2)/d*2^(1/2)+5/2*sin(d*x+c)*sec(d*x+c)^(
1/2)/a/d/(a+a*cos(d*x+c))^(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, 2984, 12, 2782, 205} \[ -\frac {7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {5 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[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]])]*Sqrt[Cos[c + d*x]]*Sq
rt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) +
(5*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[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 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 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 {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {7 a^2}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{a^3}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left (7 \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}{4 a}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (7 \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 )}{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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.51, size = 458, normalized size = 2.92 \[ \frac {2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )^{3/2} \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {4 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {5}{2};1,\frac {9}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}\right )}{70 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-35}-\frac {1}{6} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}} \csc ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}} \left (124 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-350 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+298 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-75\right )-3 \left (34 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-100 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+91 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-25\right ) \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}}\right )\right )\right )}{d (a (\cos (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[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]*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(3/2)*((4*Co
s[(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*S
qrt[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[Si
n[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))

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fricas [A]  time = 1.12, size = 136, normalized size = 0.87 \[ \frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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)^2 + 2*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)*(5*cos(d*x + c) + 4)*sin(d*x + c)/sqrt(cos(d*x
+ c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 184, normalized size = 1.17 \[ -\frac {\left (-7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+5 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-\cos \left (d x +c \right ) \sqrt {2}-4 \sqrt {2}\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}}{4 d \sin \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**(3/2)/(a*(cos(c + d*x) + 1))**(3/2), x)

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