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

[Out]

-2/5*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-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 = 0.23, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 1.62, 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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fricas [B]  time = 0.52, size = 239, normalized size = 1.74 \[ \frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + {\left (3 \, \cos \left (d x + c\right )^{3} + 14 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 52\right )} \sin \left (d x + c\right ) - 41 \, \cos \left (d x + c\right ) - 52\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{15 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

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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giac [B]  time = 0.82, size = 297, normalized size = 2.17 \[ \frac {4 \, {\left (\frac {2 \, \sqrt {2} {\left (15 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 13 \, \sqrt {-a} \sqrt {a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {-a} a^{3}} + \frac {{\left ({\left ({\left ({\left (\frac {19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {30}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {55}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {55}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {30}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {19}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}} - \frac {30 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{15 \, d} \]

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]

4/15*(2*sqrt(2)*(15*a*arctan(sqrt(a)/sqrt(-a)) + 13*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*
a^3) + (((((19*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) - 30/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*
d*x + 1/2*c) + 55/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 55/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(
1/2*d*x + 1/2*c) + 30/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 19/sgn(tan(1/2*d*x + 1/2*c) + 1))/
(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2) - 30*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(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 = 1.20, size = 130, normalized size = 0.95 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (30 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-3 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-5 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a -30 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{15 a^{5} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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