3.6.38 \(\int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx\) [538]

Optimal. Leaf size=138 \[ -\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f} \]

[Out]

-2/35*a*(7*c+5*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/7*d*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f-64/105*a^3*(7*
c+5*d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-16/105*a^2*(7*c+5*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.08, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2830, 2726, 2725} \begin {gather*} -\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*(7*c + 5*d)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(7*c + 5*d)*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]])/(105*f) - (2*a*(7*c + 5*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*f) - (2*d*Cos[e + f
*x]*(a + a*Sin[e + f*x])^(5/2))/(7*f)

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)]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx &=-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{7} (7 c+5 d) \int (a+a \sin (e+f x))^{5/2} \, dx\\ &=-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{35} (8 a (7 c+5 d)) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{105} \left (32 a^2 (7 c+5 d)\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}\\ \end {align*}

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Mathematica [A]
time = 1.01, size = 119, normalized size = 0.86 \begin {gather*} -\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (1246 c+1040 d-6 (7 c+20 d) \cos (2 (e+f x))+(392 c+505 d) \sin (e+f x)-15 d \sin (3 (e+f x)))}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/210*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(1246*c + 1040*d - 6*(7*c + 20*d)
*Cos[2*(e + f*x)] + (392*c + 505*d)*Sin[e + f*x] - 15*d*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2]))

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Maple [A]
time = 3.66, size = 99, normalized size = 0.72

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-15 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) d +\left (98 c +130 d \right ) \sin \left (f x +e \right )+\left (-21 c -60 d \right ) \left (\cos ^{2}\left (f x +e \right )\right )+322 c +290 d \right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(1+sin(f*x+e))*a^3*(sin(f*x+e)-1)*(-15*sin(f*x+e)*cos(f*x+e)^2*d+(98*c+130*d)*sin(f*x+e)+(-21*c-60*d)*co
s(f*x+e)^2+322*c+290*d)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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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((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 217, normalized size = 1.57 \begin {gather*} \frac {2 \, {\left (15 \, a^{2} d \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, a^{2} c + 20 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} - 224 \, a^{2} c - 160 \, a^{2} d - {\left (77 \, a^{2} c + 85 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (161 \, a^{2} c + 145 \, a^{2} d\right )} \cos \left (f x + e\right ) + {\left (15 \, a^{2} d \cos \left (f x + e\right )^{3} + 224 \, a^{2} c + 160 \, a^{2} d - 3 \, {\left (7 \, a^{2} c + 15 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (49 \, a^{2} c + 65 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^2*d*cos(f*x + e)^4 + 3*(7*a^2*c + 20*a^2*d)*cos(f*x + e)^3 - 224*a^2*c - 160*a^2*d - (77*a^2*c + 8
5*a^2*d)*cos(f*x + e)^2 - 2*(161*a^2*c + 145*a^2*d)*cos(f*x + e) + (15*a^2*d*cos(f*x + e)^3 + 224*a^2*c + 160*
a^2*d - 3*(7*a^2*c + 15*a^2*d)*cos(f*x + e)^2 - 2*(49*a^2*c + 65*a^2*d)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin
(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sin {\left (e + f x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 213, normalized size = 1.54 \begin {gather*} \frac {\sqrt {2} {\left (15 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 525 \, {\left (4 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (10 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 11 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (2 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*sqrt(2)*(15*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 7/2*f*x + 7/2*e) + 525*(4*a^2*c*sgn(
cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e)
+ 35*(10*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 11*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi
+ 3/2*f*x + 3/2*e) + 21*(2*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/
2*e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e))*sqrt(a)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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