3.78 \(\int \frac{\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac{9 \cos (c+d x)}{4 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(75*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (Cos[c + d*x]
*Sin[c + d*x]^2)/(4*d*(a + a*Sin[c + d*x])^(5/2)) - (13*Cos[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (9
*Cos[c + d*x])/(4*a^2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.267828, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2765, 2968, 3019, 2751, 2649, 206} \[ -\frac{9 \cos (c+d x)}{4 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(75*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (Cos[c + d*x]
*Sin[c + d*x]^2)/(4*d*(a + a*Sin[c + d*x])^(5/2)) - (13*Cos[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (9
*Cos[c + d*x])/(4*a^2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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{\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{\int \frac{\sin (c+d x) \left (2 a-\frac{9}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{\int \frac{2 a \sin (c+d x)-\frac{9}{2} a \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac{\int \frac{-\frac{39 a^2}{4}+9 a^2 \sin (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{8 a^4}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{9 \cos (c+d x)}{4 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{75 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{32 a^2}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{9 \cos (c+d x)}{4 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{75 \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 )}{16 a^2 d}\\ &=\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{9 \cos (c+d x)}{4 a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.321032, size = 173, normalized size = 1.19 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (45 \sin \left (\frac{1}{2} (c+d x)\right )-69 \sin \left (\frac{3}{2} (c+d x)\right )-16 \sin \left (\frac{5}{2} (c+d x)\right )-45 \cos \left (\frac{1}{2} (c+d x)\right )-69 \cos \left (\frac{3}{2} (c+d x)\right )+16 \cos \left (\frac{5}{2} (c+d x)\right )+(-150-150 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{32 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-45*Cos[(c + d*x)/2] - 69*Cos[(3*(c + d*x))/2] + 16*Cos[(5*(c + d*x))/
2] + 45*Sin[(c + d*x)/2] - (150 + 150*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(C
os[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 69*Sin[(3*(c + d*x))/2] - 16*Sin[(5*(c + d*x))/2]))/(32*d*(a*(1 + Sin[
c + d*x]))^(5/2))

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Maple [A]  time = 0.742, size = 233, normalized size = 1.6 \begin{align*}{\frac{1}{ \left ( 32+32\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( \sin \left ( dx+c \right ) \left ( 150\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-128\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2} \right ) + \left ( -75\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+64\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+150\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-204\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2}+42\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{a} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x)

[Out]

1/32/a^(9/2)*(sin(d*x+c)*(150*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2-128*(a-a*sin(d*x
+c))^(1/2)*a^(3/2))+(-75*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+64*(a-a*sin(d*x+c))^(
1/2)*a^(3/2))*cos(d*x+c)^2+150*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2-204*(a-a*sin(d*
x+c))^(1/2)*a^(3/2)+42*(a-a*sin(d*x+c))^(3/2)*a^(1/2))*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c))/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{\sin \left (d x + c\right )^{3}}{{\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(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [B]  time = 1.79555, size = 910, normalized size = 6.28 \begin{align*} \frac{75 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\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 ) - 4 \,{\left (32 \, \cos \left (d x + c\right )^{3} - 53 \, \cos \left (d x + c\right )^{2} -{\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) - 81 \, \cos \left (d x + c\right ) + 4\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) - 2*c
os(d*x + c) - 4)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - s
in(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x +
c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) - 4*(32*cos(d*x + c)^3 - 53*cos(d*x + c)^2 - (32*cos(d*x + c)^2 + 85
*cos(d*x + c) + 4)*sin(d*x + c) - 81*cos(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^3 + 3*a^3
*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*s
in(d*x + c))

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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(sin(d*x+c)**3/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

sage2