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

Optimal. Leaf size=184 \[ \frac{76 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{105 a^2 d}+\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac{2 \sin ^3(c+d x) \cos (c+d x)}{7 a d \sqrt{a \sin (c+d x)+a}}-\frac{16 \sin ^2(c+d x) \cos (c+d x)}{35 a d \sqrt{a \sin (c+d x)+a}}-\frac{344 \cos (c+d x)}{105 a d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(2*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(3/2)*d) - (344*Cos[c + d*x]
)/(105*a*d*Sqrt[a + a*Sin[c + d*x]]) - (16*Cos[c + d*x]*Sin[c + d*x]^2)/(35*a*d*Sqrt[a + a*Sin[c + d*x]]) + (2
*Cos[c + d*x]*Sin[c + d*x]^3)/(7*a*d*Sqrt[a + a*Sin[c + d*x]]) + (76*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(1
05*a^2*d)

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Rubi [A]  time = 0.589225, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {2879, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac{76 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{105 a^2 d}+\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac{2 \sin ^3(c+d x) \cos (c+d x)}{7 a d \sqrt{a \sin (c+d x)+a}}-\frac{16 \sin ^2(c+d x) \cos (c+d x)}{35 a d \sqrt{a \sin (c+d x)+a}}-\frac{344 \cos (c+d x)}{105 a d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^3)/(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) - (344*Cos[c + d*x]
)/(105*a*d*Sqrt[a + a*Sin[c + d*x]]) - (16*Cos[c + d*x]*Sin[c + d*x]^2)/(35*a*d*Sqrt[a + a*Sin[c + d*x]]) + (2
*Cos[c + d*x]*Sin[c + d*x]^3)/(7*a*d*Sqrt[a + a*Sin[c + d*x]]) + (76*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(1
05*a^2*d)

Rule 2879

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n]

Rule 2983

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

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

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 ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\int \frac{\sin ^3(c+d x) (a-a \sin (c+d x))}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \int \frac{\sin ^2(c+d x) \left (-3 a^2+4 a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{7 a^3}\\ &=-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{4 \int \frac{\sin (c+d x) \left (8 a^3-\frac{19}{2} a^3 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{4 \int \frac{8 a^3 \sin (c+d x)-\frac{19}{2} a^3 \sin ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{76 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 a^2 d}+\frac{8 \int \frac{-\frac{19 a^4}{4}+\frac{43}{2} a^4 \sin (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{105 a^5}\\ &=-\frac{344 \cos (c+d x)}{105 a d \sqrt{a+a \sin (c+d x)}}-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{76 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 a^2 d}-\frac{2 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{a}\\ &=-\frac{344 \cos (c+d x)}{105 a d \sqrt{a+a \sin (c+d x)}}-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{76 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 a^2 d}+\frac{4 \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 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{344 \cos (c+d x)}{105 a d \sqrt{a+a \sin (c+d x)}}-\frac{16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt{a+a \sin (c+d x)}}+\frac{76 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 a^2 d}\\ \end{align*}

Mathematica [C]  time = 1.80943, size = 201, normalized size = 1.09 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (1365 \sin \left (\frac{1}{2} (c+d x)\right )+245 \sin \left (\frac{3}{2} (c+d x)\right )-63 \sin \left (\frac{5}{2} (c+d x)\right )-15 \sin \left (\frac{7}{2} (c+d x)\right )-1365 \cos \left (\frac{1}{2} (c+d x)\right )+245 \cos \left (\frac{3}{2} (c+d x)\right )+63 \cos \left (\frac{5}{2} (c+d x)\right )-15 \cos \left (\frac{7}{2} (c+d x)\right )+(1680+1680 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\cos \left (\frac{1}{4} (2 c+d x)\right )-\sin \left (\frac{1}{4} (2 c+d x)\right )\right )\right )\right )}{420 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((1680 + 1680*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])] - 1365*Cos[(c + d*x)/2] + 245*Cos[(3*(c + d*x))/2] + 63*Cos[(5*(c + d*x))/2]
- 15*Cos[(7*(c + d*x))/2] + 1365*Sin[(c + d*x)/2] + 245*Sin[(3*(c + d*x))/2] - 63*Sin[(5*(c + d*x))/2] - 15*Si
n[(7*(c + d*x))/2]))/(420*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.967, size = 148, normalized size = 0.8 \begin{align*}{\frac{2+2\,\sin \left ( dx+c \right ) }{105\,{a}^{5}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 105\,{a}^{7/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -15\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{7/2}+21\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2}a-35\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}-105\,{a}^{3}\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)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/105*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(105*a^(7/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)
/a^(1/2))-15*(a-a*sin(d*x+c))^(7/2)+21*(a-a*sin(d*x+c))^(5/2)*a-35*(a-a*sin(d*x+c))^(3/2)*a^2-105*a^3*(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 )^{2} \sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76751, size = 725, normalized size = 3.94 \begin{align*} \frac{\frac{105 \, \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 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 24 \, \cos \left (d x + c\right )^{3} - 92 \, \cos \left (d x + c\right )^{2} +{\left (15 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} - 53 \, \cos \left (d x + c\right ) - 211\right )} \sin \left (d x + c\right ) + 158 \, \cos \left (d x + c\right ) + 211\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{105 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*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*(15*cos(d*x + c)^4 - 24*cos(d*x +
c)^3 - 92*cos(d*x + c)^2 + (15*cos(d*x + c)^3 + 39*cos(d*x + c)^2 - 53*cos(d*x + c) - 211)*sin(d*x + c) + 158*
cos(d*x + c) + 211)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*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)**2*sin(d*x+c)**3/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.39869, size = 486, normalized size = 2.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/26880*(107520*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*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2*(((((((67*sgn(tan(1/2*d*x + 1/2*c) + 1)*
tan(1/2*d*x + 1/2*c)/a^10 - 105*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) + 287*sgn(tan(1/2*d*x
+ 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) - 385*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) + 385
*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) - 287*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*
x + 1/2*c) + 105*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) - 67*sgn(tan(1/2*d*x + 1/2*c) + 1)/a
^10)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) - sqrt(2)*(107520*a^(27/2)*arctan(sqrt(a)/sqrt(-a)) + 211*sqrt(-a)*a
)*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^(29/2)))/d