3.6.22 \(\int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3 \, dx\) [522]

Optimal. Leaf size=161 \[ -\frac {4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt {a+a \sin (e+f x)}}-\frac {8 (5 c-d) d (c+d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{35 f}-\frac {12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a+a \sin (e+f x)}} \]

[Out]

-12/35*d^2*(c+d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f-4/35*a*(c+d)*(15*c^2+10*c*d+7*d^2)*cos(f*x+e)/f/(a+a*si
n(f*x+e))^(1/2)-2/7*a*cos(f*x+e)*(c+d*sin(f*x+e))^3/f/(a+a*sin(f*x+e))^(1/2)-8/35*(5*c-d)*d*(c+d)*cos(f*x+e)*(
a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2849, 2840, 2830, 2725} \begin {gather*} -\frac {4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt {a \sin (e+f x)+a}}-\frac {12 d^2 (c+d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 a f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}-\frac {8 d (5 c-d) (c+d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{35 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^3,x]

[Out]

(-4*a*(c + d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (8*(5*c - d)*d*(c + d)
*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(35*f) - (12*d^2*(c + d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*
a*f) - (2*a*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(7*f*Sqrt[a + a*Sin[e + f*x]])

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 2840

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-
d^2)*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[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], 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, -1]

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]

Rubi steps

\begin {align*} \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3 \, dx &=-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a+a \sin (e+f x)}}+\frac {1}{7} (6 (c+d)) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx\\ &=-\frac {12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a+a \sin (e+f x)}}+\frac {(12 (c+d)) \int \sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a \left (5 c^2+3 d^2\right )+a (5 c-d) d \sin (e+f x)\right ) \, dx}{35 a}\\ &=-\frac {8 (5 c-d) d (c+d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{35 f}-\frac {12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a+a \sin (e+f x)}}+\frac {1}{35} \left (2 (c+d) \left (15 c^2+10 c d+7 d^2\right )\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt {a+a \sin (e+f x)}}-\frac {8 (5 c-d) d (c+d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{35 f}-\frac {12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.34, size = 146, normalized size = 0.91 \begin {gather*} -\frac {\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))} \left (140 c^3+280 c^2 d+266 c d^2+76 d^3-6 d^2 (7 c+2 d) \cos (2 (e+f x))+d \left (140 c^2+112 c d+47 d^2\right ) \sin (e+f x)-5 d^3 \sin (3 (e+f x))\right )}{70 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[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^3,x]

[Out]

-1/70*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(140*c^3 + 280*c^2*d + 266*c*d^2 + 76*
d^3 - 6*d^2*(7*c + 2*d)*Cos[2*(e + f*x)] + d*(140*c^2 + 112*c*d + 47*d^2)*Sin[e + f*x] - 5*d^3*Sin[3*(e + f*x)
]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

________________________________________________________________________________________

Maple [A]
time = 3.20, size = 141, normalized size = 0.88

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (5 d^{3} \left (\sin ^{3}\left (f x +e \right )\right )+21 c \,d^{2} \left (\sin ^{2}\left (f x +e \right )\right )+6 d^{3} \left (\sin ^{2}\left (f x +e \right )\right )+35 c^{2} d \sin \left (f x +e \right )+28 c \,d^{2} \sin \left (f x +e \right )+8 d^{3} \sin \left (f x +e \right )+35 c^{3}+70 c^{2} d +56 c \,d^{2}+16 d^{3}\right )}{35 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*(5*d^3*sin(f*x+e)^3+21*c*d^2*sin(f*x+e)^2+6*d^3*sin(f*x+e)^2+35*c^2*d*sin
(f*x+e)+28*c*d^2*sin(f*x+e)+8*d^3*sin(f*x+e)+35*c^3+70*c^2*d+56*c*d^2+16*d^3)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2
)/f

________________________________________________________________________________________

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))^(1/2)*(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 251, normalized size = 1.56 \begin {gather*} \frac {2 \, {\left (5 \, d^{3} \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 35 \, c^{3} - 35 \, c^{2} d - 49 \, c d^{2} - 9 \, d^{3} - {\left (35 \, c^{2} d + 7 \, c d^{2} + 12 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, c^{3} + 70 \, c^{2} d + 77 \, c d^{2} + 22 \, d^{3}\right )} \cos \left (f x + e\right ) + {\left (5 \, d^{3} \cos \left (f x + e\right )^{3} + 35 \, c^{3} + 35 \, c^{2} d + 49 \, c d^{2} + 9 \, d^{3} - {\left (21 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, c^{2} d + 28 \, c d^{2} + 13 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{35 \, {\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))^(1/2)*(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

2/35*(5*d^3*cos(f*x + e)^4 + 3*(7*c*d^2 + 2*d^3)*cos(f*x + e)^3 - 35*c^3 - 35*c^2*d - 49*c*d^2 - 9*d^3 - (35*c
^2*d + 7*c*d^2 + 12*d^3)*cos(f*x + e)^2 - (35*c^3 + 70*c^2*d + 77*c*d^2 + 22*d^3)*cos(f*x + e) + (5*d^3*cos(f*
x + e)^3 + 35*c^3 + 35*c^2*d + 49*c*d^2 + 9*d^3 - (21*c*d^2 + d^3)*cos(f*x + e)^2 - (35*c^2*d + 28*c*d^2 + 13*
d^3)*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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(1/2)*(c+d*sin(f*x+e))**3,x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(c + d*sin(e + f*x))**3, x)

________________________________________________________________________________________

Giac [A]
time = 0.60, size = 266, normalized size = 1.65 \begin {gather*} \frac {\sqrt {2} {\left (5 \, d^{3} \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 ) + 35 \, {\left (8 \, c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, d^{3} \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 (4 \, c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{3} \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 ) + 7 \, {\left (6 \, c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{3} \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}}{140 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/140*sqrt(2)*(5*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 7/2*f*x + 7/2*e) + 35*(8*c^3*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)) + 12*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)) + 3*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 35*(4*c^2*d*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)) + 2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*
e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e) + 7*(6*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^3*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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________