3.1.78 \(\int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [78]

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

[Out]

1/4*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(5/2)-13/16*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)+75/32*arctanh
(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-9/4*cos(d*x+c)/a^2/d/(a+a*sin(d*x+c)
)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 145, normalized size of antiderivative = 1.00, 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 = {2844, 3047, 3098, 2830, 2728, 212} \begin {gather*} \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 {9 \cos (c+d x)}{4 a^2 d \sqrt {a \sin (c+d x)+a}}+\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}} \end {gather*}

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 212

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

Rule 2728

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 2830

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*S
in[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 2844

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 3047

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 3098

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]

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 \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.22, size = 173, normalized size = 1.19 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-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 )+45 \sin \left (\frac {1}{2} (c+d x)\right )-(150+150 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-69 \sin \left (\frac {3}{2} (c+d x)\right )-16 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{32 d (a (1+\sin (c+d x)))^{5/2}} \end {gather*}

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 = 2.68, size = 233, normalized size = 1.61

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \left (150 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-128 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right )+\left (-75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+64 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+150 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+42 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-204 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(233\)

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,method=_RETURNVERBOSE)

[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+42*(a-a*sin(d*x
+c))^(3/2)*a^(1/2)-204*(a-a*sin(d*x+c))^(1/2)*a^(3/2))*(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (122) = 244\).
time = 0.35, size = 341, normalized size = 2.35 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 [A]
time = 0.49, size = 191, normalized size = 1.32 \begin {gather*} -\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {128 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (21 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 19 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \end {gather*}

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]

-1/64*(75*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 75*s
qrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 128*sqrt(2)*si
n(-1/4*pi + 1/2*d*x + 1/2*c)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*sqrt(2)*(21*sin(-1/4*pi + 1/2*d
*x + 1/2*c)^3 - 19*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^(5/2)*sgn(cos(-
1/4*pi + 1/2*d*x + 1/2*c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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