3.188 \(\int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac {4 \cos (c+d x)}{a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2679, 2649, 206} \[ \frac {4 \cos (c+d x)}{a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

Rule 206

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

Rule 2649

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}+\frac {2 \int \frac {\cos ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{a}\\ &=\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}+\frac {4 \cos (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}+\frac {4 \cos (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {8 \operatorname {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^2 d}\\ &=-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}+\frac {4 \cos (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 96, normalized size = 0.89 \[ -\frac {2 \cos (c+d x) \left (\sqrt {1-\sin (c+d x)} (\sin (c+d x)-7)+6 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )\right )}{3 a^2 d \sqrt {1-\sin (c+d x)} \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*Cos[c + d*x]*(6*Sqrt[2]*ArcTanh[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]] + Sqrt[1 - Sin[c + d*x]]*(-7 + Sin[c + d*x
])))/(3*a^2*d*Sqrt[1 - Sin[c + d*x]]*Sqrt[a*(1 + Sin[c + d*x])])

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fricas [B]  time = 0.68, size = 215, normalized size = 1.99 \[ \frac {2 \, {\left (\frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right ) - 7 \, \cos \left (d x + c\right ) - 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{3 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*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) - (cos(d*x + c)^2 + (cos(d*x + c) + 8)*sin
(d*x + c) - 7*cos(d*x + c) - 8)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

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giac [B]  time = 3.12, size = 255, normalized size = 2.36 \[ -\frac {2 \, {\left (\frac {{\left ({\left (\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {9}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {9}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {7}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {4 \, \sqrt {2} {\left (3 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 2 \, \sqrt {-a} \sqrt {a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {-a} a^{3}} - \frac {12 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-2/3*((((7*tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 9/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1
/2*d*x + 1/2*c) + 9/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) - 7/(a*sgn(tan(1/2*d*x + 1/2*c) +
1)))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) + 4*sqrt(2)*(3*a*arctan(sqrt(a)/sqrt(-a)) + 2*sqrt(-a)*sqrt(a))*sgn(
tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^3) - 12*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt
(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.23, size = 112, normalized size = 1.04 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (6 a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-6 a \sqrt {a -a \sin \left (d x +c \right )}\right )}{3 a^{4} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(6*a^(3/2)*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))^(3/2)-6*a*(a-a*sin(d*x+c))^(1/2))/a^4/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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^(5/2), 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(cos(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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