3.3.74 \(\int \frac {\sqrt {1-\cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [274]

Optimal. Leaf size=112 \[ \frac {2 \sin (c+d x)}{5 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)}-\frac {8 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}} \]

[Out]

2/5*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(1-cos(d*x+c))^(1/2)-8/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(1-cos(d*x+c))^(1/2)
+16/15*sin(d*x+c)/d/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2851, 2850} \begin {gather*} -\frac {8 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{5 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sin[c + d*x])/(5*d*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)) - (8*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x
]]*Cos[c + d*x]^(3/2)) + (16*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*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 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*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[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
 + 1), 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] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {\sqrt {1-\cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 \sin (c+d x)}{5 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{5} \int \frac {\sqrt {1-\cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)}-\frac {8 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {8}{15} \int \frac {\sqrt {1-\cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)}-\frac {8 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.12, size = 61, normalized size = 0.54 \begin {gather*} \frac {2 \sqrt {1-\cos (c+d x)} \left (3-4 \cos (c+d x)+8 \cos ^2(c+d x)\right ) \cot \left (\frac {1}{2} (c+d x)\right )}{15 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - Cos[c + d*x]]*(3 - 4*Cos[c + d*x] + 8*Cos[c + d*x]^2)*Cot[(c + d*x)/2])/(15*d*Cos[c + d*x]^(5/2))

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 65, normalized size = 0.58

method result size
default \(-\frac {\left (8 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )+3\right ) \sqrt {2-2 \cos \left (d x +c \right )}\, \sin \left (d x +c \right ) \sqrt {2}}{15 d \left (-1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\) \(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/d*(8*cos(d*x+c)^2-4*cos(d*x+c)+3)*(2-2*cos(d*x+c))^(1/2)*sin(d*x+c)/(-1+cos(d*x+c))/cos(d*x+c)^(5/2)*2^(
1/2)

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (94) = 188\).
time = 0.53, size = 209, normalized size = 1.87 \begin {gather*} \frac {2 \, {\left (7 \, \sqrt {2} - \frac {17 \, \sqrt {2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {15 \, \sqrt {2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(7*sqrt(2) - 17*sqrt(2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 25*sqrt(2)*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 15*sqrt(2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/(d*(sin(d*x
+ c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(cos(d*x + c
) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1))

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 63, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left (8 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {-\cos \left (d x + c\right ) + 1}}{15 \, d \cos \left (d x + c\right )^{\frac {5}{2}} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(8*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(-cos(d*x + c) + 1)/(d*cos(d*x + c)^(5/2)*si
n(d*x + c))

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c))**(1/2)/cos(d*x+c)**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

________________________________________________________________________________________

Giac [A]
time = 0.61, size = 117, normalized size = 1.04 \begin {gather*} -\frac {2 \, \sqrt {2} {\left ({\left ({\left ({\left ({\left (7 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 75\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 430\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 430\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 75\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 7\right )} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{15 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*sqrt(2)*(((((7*tan(1/4*d*x + 1/4*c)^2 - 75)*tan(1/4*d*x + 1/4*c)^2 + 430)*tan(1/4*d*x + 1/4*c)^2 - 430)*
tan(1/4*d*x + 1/4*c)^2 + 75)*tan(1/4*d*x + 1/4*c)^2 - 7)*sgn(sin(1/2*d*x + 1/2*c))/((tan(1/4*d*x + 1/4*c)^4 -
6*tan(1/4*d*x + 1/4*c)^2 + 1)^(5/2)*d)

________________________________________________________________________________________

Mupad [B]
time = 2.00, size = 156, normalized size = 1.39 \begin {gather*} \frac {8\,\sqrt {2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\,\left (7\,\sin \left (c+d\,x\right )-4\,\sin \left (2\,c+2\,d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )-2\,\sin \left (4\,c+4\,d\,x\right )+2\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\sqrt {1-2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\,\left (-16\,{\sin \left (c+d\,x\right )}^2-4\,{\sin \left (2\,c+2\,d\,x\right )}^2+20\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+10\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+2\,{\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(2*sin(c/2 + (d*x)/2)^2)^(1/2)*(7*sin(c + d*x) - 4*sin(2*c + 2*d*x) + 9*sin(3*c + 3*d*x) - 2*sin(4*c + 4*d*
x) + 2*sin(5*c + 5*d*x)))/(15*d*(1 - 2*sin(c/2 + (d*x)/2)^2)^(1/2)*(20*sin(c/2 + (d*x)/2)^2 - 4*sin(2*c + 2*d*
x)^2 + 10*sin((3*c)/2 + (3*d*x)/2)^2 + 2*sin((5*c)/2 + (5*d*x)/2)^2 - 16*sin(c + d*x)^2))

________________________________________________________________________________________