3.281 \(\int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=173 \[ \frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

-arctanh(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))*2^(1/2)/d/a^(1/2)+2/5*sin(d*x
+c)/d/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2)+2/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a-a*cos(d*x+c))^(1/2)+26/15*
sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2779, 2984, 12, 2782, 208} \[ \frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*Sqrt[a - a*Cos[c + d*x]]),x]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/(Sqrt[a]*d))
+ (2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]) + (2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3
/2)*Sqrt[a - a*Cos[c + d*x]]) + (26*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*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 208

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

Rule 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e +
f*x], x])/Sqrt[a + b*Sin[e + f*x]], 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] && LtQ[n, -1] && IntegerQ[2*n]

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])

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {\int \frac {a+4 a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {2 \int \frac {-\frac {13 a^2}{2}-a^2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}+\frac {4 \int \frac {15 a^3}{4 \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \, dx}{15 a^3}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}+\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {(2 a) \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 )}{d}\\ &=-\frac {\sqrt {2} \tanh ^{-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 {a} d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.66, size = 218, normalized size = 1.26 \[ \frac {e^{-\frac {5}{2} i (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right ) \left (2 \sqrt {1+e^{2 i (c+d x)}} \left (15 e^{i (c+d x)}+40 e^{2 i (c+d x)}+40 e^{3 i (c+d x)}+15 e^{4 i (c+d x)}+13 e^{5 i (c+d x)}+13\right )-15 \sqrt {2} \left (1+e^{2 i (c+d x)}\right )^3 \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{60 d \sqrt {1+e^{2 i (c+d x)}} \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*Sqrt[a - a*Cos[c + d*x]]),x]

[Out]

((2*Sqrt[1 + E^((2*I)*(c + d*x))]*(13 + 15*E^(I*(c + d*x)) + 40*E^((2*I)*(c + d*x)) + 40*E^((3*I)*(c + d*x)) +
15*E^((4*I)*(c + d*x)) + 13*E^((5*I)*(c + d*x))) - 15*Sqrt[2]*(1 + E^((2*I)*(c + d*x)))^3*ArcTanh[(1 + E^(I*(
c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])*Sin[(c + d*x)/2])/(60*d*E^(((5*I)/2)*(c + d*x))*Sqrt[1 +
E^((2*I)*(c + d*x))]*Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]])

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fricas [A]  time = 0.91, size = 177, normalized size = 1.02 \[ \frac {15 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (13 \, \cos \left (d x + c\right )^{3} + 14 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{30 \, a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30*(15*sqrt(2)*sqrt(a)*cos(d*x + c)^3*log(-(2*sqrt(2)*sqrt(-a*cos(d*x + c) + a)*(cos(d*x + c) + 1)*sqrt(cos(
d*x + c))/sqrt(a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) + 4*(13
*cos(d*x + c)^3 + 14*cos(d*x + c)^2 + 4*cos(d*x + c) + 3)*sqrt(-a*cos(d*x + c) + a)*sqrt(cos(d*x + c)))/(a*d*c
os(d*x + c)^3*sin(d*x + c))

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giac [A]  time = 1.28, size = 136, normalized size = 0.79 \[ \frac {\sqrt {2} a {\left (\frac {15 \, \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} {\left | a \right |}} + \frac {2 \, {\left (15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} + 10 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 12 \, a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left | a \right |}}\right )}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/15*sqrt(2)*a*(15*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*abs(a)) + 2*(15*(a*tan(1/2*d
*x + 1/2*c)^2 - a)^2 + 10*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a + 12*a^2)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a
*tan(1/2*d*x + 1/2*c)^2 + a)*abs(a)))/d

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maple [B]  time = 0.20, size = 305, normalized size = 1.76 \[ \frac {\left (\sin ^{7}\left (d x +c \right )\right ) \left (15 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+45 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+45 \sqrt {2}\, \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+15 \sqrt {2}\, \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-26 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-6 \cos \left (d x +c \right )\right ) \sqrt {2}}{15 d \left (-1+\cos \left (d x +c \right )\right )^{3} \sqrt {-2 a \left (-1+\cos \left (d x +c \right )\right )}\, \left (1+\cos \left (d x +c \right )\right )^{3} \cos \left (d x +c \right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(7/2)/(a-a*cos(d*x+c))^(1/2),x)

[Out]

1/15/d*sin(d*x+c)^7*(15*2^(1/2)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*2^(1/2)/(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2))+45*2^(1/2)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*2^(1/2)/(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2))+45*2^(1/2)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*2^(1/2)/(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2))+15*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*2^(1/2)/(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2))-26*cos(d*x+c)^3-2*cos(d*x+c)^2-6*cos(d*x+c))/(-1+cos(d*x+c))^3/(-2*a*(-1+cos(d*x+c)))^(1/2)/
(1+cos(d*x+c))^3/cos(d*x+c)^(7/2)*2^(1/2)

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maxima [C]  time = 0.98, size = 692, normalized size = 4.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/15*(15*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*(co
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*arctan2(2*sqrt(2)*(cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))/(s
qrt(a)*abs(e^(I*d*x + I*c) - 1)), 2*(sqrt(2)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)^(1/4)*sqrt(a)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt(-a)*abs(e^(I*d*x + I*c) - 1) +
2*sqrt(a))/(a*abs(e^(I*d*x + I*c) - 1))) - 26*(cos(2*d*x + 2*c)^2*sin(d*x + c) + sin(2*d*x + 2*c)^2*sin(d*x +
c) + 2*cos(2*d*x + 2*c)*sin(d*x + c) + sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
+ 24*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 24*(cos(d*x + c) + 1)*sin(5/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*((13*cos(d*x + c) + 15)*cos(2*d*x + 2*c)^2 + (13*cos(d*x +
c) + 15)*sin(2*d*x + 2*c)^2 + 2*(13*cos(d*x + c) + 15)*cos(2*d*x + 2*c) + 13*cos(d*x + c) + 15)*sin(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*
c) + 1)*(7*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (7*cos(d*x + c) + 5)*sin(3/
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(5/4)*sqrt(-a)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(7/2)*(a - a*cos(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^(7/2)*(a - a*cos(c + d*x))^(1/2)), 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(1/cos(d*x+c)**(7/2)/(a-a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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