3.1.61 \(\int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [61]

Optimal. Leaf size=139 \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {28 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d} \]

[Out]

arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/d*2^(1/2)/a^(1/2)-28/15*cos(d*x+c)/d/(a+a*sin(d
*x+c))^(1/2)-2/5*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)+2/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a/d

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Rubi [A]
time = 0.16, antiderivative size = 139, 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 = {2857, 3047, 3102, 2830, 2728, 212} \begin {gather*} -\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 a d}-\frac {28 \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[a]*d) - (28*Cos[c + d*x])/(
15*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*Sqrt[a + a*Sin[c + d*x]]) + (2*Cos[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(15*a*d)

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 2857

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n -
1)) + d*(a*d - b*c*(4*n - 3))*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] && GtQ[n, 1] && IntegerQ[2*n]

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 3102

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]

Rubi steps

\begin {align*} \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {\sin (c+d x) (-4 a+a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {-4 a \sin (c+d x)+a \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}-\frac {2 \int \frac {\frac {a^2}{2}-7 a^2 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {28 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}-\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {28 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}+\frac {2 \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 )}{d}\\ &=\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {28 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.16, size = 150, normalized size = 1.08 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left ((-60-60 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 )-60 \cos \left (\frac {1}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )+3 \cos \left (\frac {5}{2} (c+d x)\right )+60 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \sqrt {a (1+\sin (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*((-60 - 60*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c +
d*x)/4])] - 60*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x))/2] + 3*Cos[(5*(c + d*x))/2] + 60*Sin[(c + d*x)/2] + 5*Si
n[(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2]))/(30*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]
time = 2.20, size = 130, normalized size = 0.94

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-6 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}+10 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a -30 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{15 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(15*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a
^(1/2))-6*(a-a*sin(d*x+c))^(5/2)+10*(a-a*sin(d*x+c))^(3/2)*a-30*a^2*(a-a*sin(d*x+c))^(1/2))/a^3/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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 234, normalized size = 1.68 \begin {gather*} \frac {\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}} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 17\right )} \sin \left (d x + c\right ) - 16 \, \cos \left (d x + c\right ) - 17\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\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))^(1/2),x, algorithm="fricas")

[Out]

1/30*(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) + 4*(3*cos(d*x + c)^3 + 4*cos(d*x + c)^2
- (3*cos(d*x + c)^2 - cos(d*x + c) - 17)*sin(d*x + c) - 16*cos(d*x + c) - 17)*sqrt(a*sin(d*x + c) + a))/(a*d*
cos(d*x + c) + a*d*sin(d*x + c) + a*d)

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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))**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 0.53, size = 163, normalized size = 1.17 \begin {gather*} -\frac {\frac {15 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (12 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{30 \, 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))^(1/2),x, algorithm="giac")

[Out]

-1/30*(15*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 15*s
qrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 4*sqrt(2)*(12*
a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 10*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 15*a^(9/2)*sin(-1/4*p
i + 1/2*d*x + 1/2*c))/(a^5*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}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,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))^(1/2),x)

[Out]

int(sin(c + d*x)^3/(a + a*sin(c + d*x))^(1/2), x)

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