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

Optimal. Leaf size=137 \[ -\frac{4 \cos (c+d x)}{a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

[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]^5
)/(5*d*(a + a*Sin[c + d*x])^(5/2)) - (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]])

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Rubi [A]  time = 0.23456, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2860, 2679, 2649, 206} \[ -\frac{4 \cos (c+d x)}{a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(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]^5
)/(5*d*(a + a*Sin[c + d*x])^(5/2)) - (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 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 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]

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 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])

Rubi steps

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

Mathematica [C]  time = 1.49299, size = 177, normalized size = 1.29 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (180 \sin \left (\frac{1}{2} (c+d x)\right )+25 \sin \left (\frac{3}{2} (c+d x)\right )-3 \sin \left (\frac{5}{2} (c+d x)\right )-180 \cos \left (\frac{1}{2} (c+d x)\right )+25 \cos \left (\frac{3}{2} (c+d x)\right )+3 \cos \left (\frac{5}{2} (c+d x)\right )+(240+240 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\cos \left (\frac{1}{4} (2 c+d x)\right )-\sin \left (\frac{1}{4} (2 c+d x)\right )\right )\right )\right )}{30 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((240 + 240*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c +
d*x)/4] - Sin[(2*c + d*x)/4])] - 180*Cos[(c + d*x)/2] + 25*Cos[(3*(c + d*x))/2] + 3*Cos[(5*(c + d*x))/2] + 180
*Sin[(c + d*x)/2] + 25*Sin[(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2]))/(30*a^3*d*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]))

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Maple [A]  time = 0.844, size = 130, normalized size = 1. \begin{align*}{\frac{2+2\,\sin \left ( dx+c \right ) }{15\,{a}^{5}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 30\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -3\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2}-5\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}a-30\,{a}^{2}\sqrt{a-a\sin \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(30*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a
^(1/2))-3*(a-a*sin(d*x+c))^(5/2)-5*(a-a*sin(d*x+c))^(3/2)*a-30*a^2*(a-a*sin(d*x+c))^(1/2))/a^5/cos(d*x+c)/(a+a
*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.11816, size = 659, normalized size = 4.81 \begin{align*} \frac{2 \,{\left (\frac{15 \, \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 (3 \, \cos \left (d x + c\right )^{3} + 14 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 52\right )} \sin \left (d x + c\right ) - 41 \, \cos \left (d x + c\right ) - 52\right )} \sqrt{a \sin \left (d x + c\right ) + a}\right )}}{15 \,{\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*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) + (3*cos(d*x + c)^3 + 14*cos(d*x + c)^2
- (3*cos(d*x + c)^2 - 11*cos(d*x + c) - 52)*sin(d*x + c) - 41*cos(d*x + c) - 52)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.41112, size = 409, normalized size = 2.99 \begin{align*} \frac{\frac{{\left ({\left ({\left ({\left (\frac{19 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{9}} - \frac{30 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{55 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{55 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{30 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{19 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}} - \frac{240 \, \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 )} + \frac{2 \, \sqrt{2}{\left (120 \, a^{\frac{21}{2}} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + 13 \, \sqrt{-a} a\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{-a} a^{\frac{25}{2}}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*((((((19*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^9 - 30*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*t
an(1/2*d*x + 1/2*c) + 55*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 55*sgn(tan(1/2*d*x + 1/2*c)
+ 1)/a^9)*tan(1/2*d*x + 1/2*c) + 30*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 19*sgn(tan(1/2*
d*x + 1/2*c) + 1)/a^9)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2) - 240*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)) + 2*sqrt(2)*(120*a^(21/2)*arctan(sqrt(a)/sqrt(-a)) + 13*sqrt(-a)*a)*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt
(-a)*a^(25/2)))/d