3.2.74 \(\int \frac {\cos ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [174]

Optimal. Leaf size=76 \[ -\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {2 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2758, 2728, 212} \begin {gather*} \frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {2 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a}\\ &=\frac {2 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a d}\\ &=-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {2 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 84, normalized size = 1.11 \begin {gather*} \frac {2 \cos ^3(c+d x) \left (-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {1-\sin (c+d x)}\right )}{d (1-\sin (c+d x))^{3/2} (a (1+\sin (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^3*(-(Sqrt[2]*ArcTanh[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]) + Sqrt[1 - Sin[c + d*x]]))/(d*(1 - Sin[c
+ d*x])^(3/2)*(a*(1 + Sin[c + d*x]))^(3/2))

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Maple [A]
time = 0.54, size = 95, normalized size = 1.25

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-\sqrt {a -a \sin \left (d x +c \right )}\right )}{a^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (65) = 130\).
time = 0.37, size = 196, normalized size = 2.58 \begin {gather*} \frac {\frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) - 2*sqrt
(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 -
(cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) + 2*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin
(d*x + c) + 1))/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.22, size = 114, normalized size = 1.50 \begin {gather*} \frac {\sqrt {2} \sqrt {a} {\left (\frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*sqrt(a)*(log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - log(-sin(
-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(
a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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