3.1.53 \(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [53]

Optimal. Leaf size=146 \[ -\frac {832 a^3 \cos (c+d x)}{315 d \sqrt {a+a \sin (c+d x)}}-\frac {208 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d} \]

[Out]

-26/105*a*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d+4/63*cos(d*x+c)*(a+a*sin(d*x+c))^(5/2)/d-2/9*cos(d*x+c)*(a+a*sin
(d*x+c))^(7/2)/a/d-832/315*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-208/315*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/
2)/d

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Rubi [A]
time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2726, 2725} \begin {gather*} -\frac {832 a^3 \cos (c+d x)}{315 d \sqrt {a \sin (c+d x)+a}}-\frac {208 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{7/2}}{9 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{63 d}-\frac {26 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-832*a^3*Cos[c + d*x])/(315*d*Sqrt[a + a*Sin[c + d*x]]) - (208*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(31
5*d) - (26*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(6
3*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/2))/(9*a*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 2726

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[a*((2*n - 1)/n), Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 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)]

Rubi steps

\begin {align*} \int \sin ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac {2 \int \left (\frac {7 a}{2}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{5/2} \, dx}{9 a}\\ &=\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac {13}{21} \int (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac {26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac {1}{105} (104 a) \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {208 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac {1}{315} \left (416 a^2\right ) \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {832 a^3 \cos (c+d x)}{315 d \sqrt {a+a \sin (c+d x)}}-\frac {208 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}\\ \end {align*}

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Mathematica [A]
time = 0.66, size = 165, normalized size = 1.13 \begin {gather*} \frac {(a (1+\sin (c+d x)))^{5/2} \left (-8190 \cos \left (\frac {1}{2} (c+d x)\right )-2100 \cos \left (\frac {3}{2} (c+d x)\right )+756 \cos \left (\frac {5}{2} (c+d x)\right )+225 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )+8190 \sin \left (\frac {1}{2} (c+d x)\right )-2100 \sin \left (\frac {3}{2} (c+d x)\right )-756 \sin \left (\frac {5}{2} (c+d x)\right )+225 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 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]^2*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((a*(1 + Sin[c + d*x]))^(5/2)*(-8190*Cos[(c + d*x)/2] - 2100*Cos[(3*(c + d*x))/2] + 756*Cos[(5*(c + d*x))/2] +
225*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 8190*Sin[(c + d*x)/2] - 2100*Sin[(3*(c + d*x))/2] - 756*
Sin[(5*(c + d*x))/2] + 225*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^5)

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Maple [A]
time = 1.77, size = 85, normalized size = 0.58

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (35 \left (\sin ^{4}\left (d x +c \right )\right )+130 \left (\sin ^{3}\left (d x +c \right )\right )+219 \left (\sin ^{2}\left (d x +c \right )\right )+292 \sin \left (d x +c \right )+584\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/315*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+130*sin(d*x+c)^3+219*sin(d*x+c)^2+292*sin(d*x+c)+584)
/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)^2*(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)^2, x)

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Fricas [A]
time = 0.38, size = 167, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{5} - 95 \, a^{2} \cos \left (d x + c\right )^{4} - 289 \, a^{2} \cos \left (d x + c\right )^{3} + 263 \, a^{2} \cos \left (d x + c\right )^{2} + 838 \, a^{2} \cos \left (d x + c\right ) + 416 \, a^{2} - {\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} - 159 \, a^{2} \cos \left (d x + c\right )^{2} - 422 \, a^{2} \cos \left (d x + c\right ) + 416 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\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)^2*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/315*(35*a^2*cos(d*x + c)^5 - 95*a^2*cos(d*x + c)^4 - 289*a^2*cos(d*x + c)^3 + 263*a^2*cos(d*x + c)^2 + 838*
a^2*cos(d*x + c) + 416*a^2 - (35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 - 159*a^2*cos(d*x + c)^2 - 422*a^
2*cos(d*x + c) + 416*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \sin ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**(5/2)*sin(c + d*x)**2, x)

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Giac [A]
time = 0.62, size = 162, normalized size = 1.11 \begin {gather*} \frac {\sqrt {2} {\left (8190 \, 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 ) + 2100 \, 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 ) + 756 \, 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 ) + 225 \, 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 ) + 35 \, 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 )\right )} \sqrt {a}}{2520 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(8190*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) + 2100*a^2*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 756*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(
-5/4*pi + 5/2*d*x + 5/2*c) + 225*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) + 35*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))*sqrt(a)/d

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

[Out]

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

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