3.1.52 \(\int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [52]

Optimal. Leaf size=203 \[ -\frac {284 a^3 \cos (c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {568 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 d}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d} \]

[Out]

-284/231*a*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-284/99*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-710/693*a^3*cos(
d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-46/99*a^3*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+568/69
3*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-2/11*a^2*cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.24, antiderivative size = 203, 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 = {2842, 3060, 2849, 2838, 2830, 2725} \begin {gather*} -\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {710 a^3 \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt {a \sin (c+d x)+a}}-\frac {284 a^3 \cos (c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}+\frac {568 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{693 d}-\frac {284 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{231 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-284*a^3*Cos[c + d*x])/(99*d*Sqrt[a + a*Sin[c + d*x]]) - (710*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(693*d*Sqrt[a
+ a*Sin[c + d*x]]) - (46*a^3*Cos[c + d*x]*Sin[c + d*x]^4)/(99*d*Sqrt[a + a*Sin[c + d*x]]) + (568*a^2*Cos[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(693*d) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) -
(284*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(231*d)

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*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 2838

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

Rule 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/(
d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d
*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m,
2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2849

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[2*n*((b*c + a*d)
/(b*(2*n + 1))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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, 0] && IntegerQ[2*n]

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {2}{11} \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {19 a^2}{2}+\frac {23}{2} a^2 \sin (c+d x)\right ) \, dx\\ &=-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {1}{99} \left (355 a^2\right ) \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {1}{231} \left (710 a^2\right ) \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}+\frac {1}{231} (284 a) \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {568 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 d}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}+\frac {1}{99} \left (142 a^2\right ) \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {284 a^3 \cos (c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {568 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 d}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}\\ \end {align*}

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Mathematica [A]
time = 0.80, size = 189, normalized size = 0.93 \begin {gather*} -\frac {(a (1+\sin (c+d x)))^{5/2} \left (31878 \cos \left (\frac {1}{2} (c+d x)\right )+8778 \cos \left (\frac {3}{2} (c+d x)\right )-3465 \cos \left (\frac {5}{2} (c+d x)\right )-1287 \cos \left (\frac {7}{2} (c+d x)\right )+385 \cos \left (\frac {9}{2} (c+d x)\right )+63 \cos \left (\frac {11}{2} (c+d x)\right )-31878 \sin \left (\frac {1}{2} (c+d x)\right )+8778 \sin \left (\frac {3}{2} (c+d x)\right )+3465 \sin \left (\frac {5}{2} (c+d x)\right )-1287 \sin \left (\frac {7}{2} (c+d x)\right )-385 \sin \left (\frac {9}{2} (c+d x)\right )+63 \sin \left (\frac {11}{2} (c+d x)\right )\right )}{11088 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/11088*((a*(1 + Sin[c + d*x]))^(5/2)*(31878*Cos[(c + d*x)/2] + 8778*Cos[(3*(c + d*x))/2] - 3465*Cos[(5*(c +
d*x))/2] - 1287*Cos[(7*(c + d*x))/2] + 385*Cos[(9*(c + d*x))/2] + 63*Cos[(11*(c + d*x))/2] - 31878*Sin[(c + d*
x)/2] + 8778*Sin[(3*(c + d*x))/2] + 3465*Sin[(5*(c + d*x))/2] - 1287*Sin[(7*(c + d*x))/2] - 385*Sin[(9*(c + d*
x))/2] + 63*Sin[(11*(c + d*x))/2]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]
time = 1.86, size = 95, normalized size = 0.47

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (63 \left (\sin ^{5}\left (d x +c \right )\right )+224 \left (\sin ^{4}\left (d x +c \right )\right )+355 \left (\sin ^{3}\left (d x +c \right )\right )+426 \left (\sin ^{2}\left (d x +c \right )\right )+568 \sin \left (d x +c \right )+1136\right )}{693 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(95\)

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]

2/693*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(63*sin(d*x+c)^5+224*sin(d*x+c)^4+355*sin(d*x+c)^3+426*sin(d*x+c)^2+56
8*sin(d*x+c)+1136)/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((a*sin(d*x + c) + a)^(5/2)*sin(d*x + c)^3, x)

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Fricas [A]
time = 0.33, size = 192, normalized size = 0.95 \begin {gather*} -\frac {2 \, {\left (63 \, a^{2} \cos \left (d x + c\right )^{6} + 224 \, a^{2} \cos \left (d x + c\right )^{5} - 320 \, a^{2} \cos \left (d x + c\right )^{4} - 874 \, a^{2} \cos \left (d x + c\right )^{3} + 593 \, a^{2} \cos \left (d x + c\right )^{2} + 1786 \, a^{2} \cos \left (d x + c\right ) + 800 \, a^{2} + {\left (63 \, a^{2} \cos \left (d x + c\right )^{5} - 161 \, a^{2} \cos \left (d x + c\right )^{4} - 481 \, a^{2} \cos \left (d x + c\right )^{3} + 393 \, a^{2} \cos \left (d x + c\right )^{2} + 986 \, a^{2} \cos \left (d x + c\right ) - 800 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{693 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + 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))^(5/2),x, algorithm="fricas")

[Out]

-2/693*(63*a^2*cos(d*x + c)^6 + 224*a^2*cos(d*x + c)^5 - 320*a^2*cos(d*x + c)^4 - 874*a^2*cos(d*x + c)^3 + 593
*a^2*cos(d*x + c)^2 + 1786*a^2*cos(d*x + c) + 800*a^2 + (63*a^2*cos(d*x + c)^5 - 161*a^2*cos(d*x + c)^4 - 481*
a^2*cos(d*x + c)^3 + 393*a^2*cos(d*x + c)^2 + 986*a^2*cos(d*x + c) - 800*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c
) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + 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))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.50, size = 192, normalized size = 0.95 \begin {gather*} \frac {\sqrt {2} {\left (31878 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8778 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3465 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1287 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 385 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 63 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )} \sqrt {a}}{11088 \, 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]

1/11088*sqrt(2)*(31878*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 8778*a^2*sgn(c
os(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 3465*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s
in(-5/4*pi + 5/2*d*x + 5/2*c) + 1287*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) +
385*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c) + 63*a^2*sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c))*sin(-11/4*pi + 11/2*d*x + 11/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\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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