\(\int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx\) [303]

Optimal result

Integrand size = 25, antiderivative size = 138 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=-\frac {64 a^3 (7 A+5 B) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2830, 2726, 2725} \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=-\frac {64 a^3 (7 A+5 B) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 (7 A+5 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {2 a (7 A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f} \]

[In]

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

[Out]

(-64*a^3*(7*A + 5*B)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(7*A + 5*B)*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]])/(105*f) - (2*a*(7*A + 5*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*f) - (2*B*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*} \text {integral}& = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{7} (7 A+5 B) \int (a+a \sin (e+f x))^{5/2} \, dx \\ & = -\frac {2 a (7 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{35} (8 a (7 A+5 B)) \int (a+a \sin (e+f x))^{3/2} \, dx \\ & = -\frac {16 a^2 (7 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{105} \left (32 a^2 (7 A+5 B)\right ) \int \sqrt {a+a \sin (e+f x)} \, dx \\ & = -\frac {64 a^3 (7 A+5 B) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.56 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (1246 A+1040 B-6 (7 A+20 B) \cos (2 (e+f x))+(392 A+505 B) \sin (e+f x)-15 B \sin (3 (e+f x)))}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*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*A + 1040*B - 6*(7*A + 20*B)
*Cos[2*(e + f*x)] + (392*A + 505*B)*Sin[e + f*x] - 15*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2]))

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-15 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-21 A -60 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (98 A +130 B \right ) \sin \left (f x +e \right )+322 A +290 B \right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(99\)
parts	\(\frac {2 A \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )+14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )+12 \left (\sin ^{2}\left (f x +e \right )\right )+23 \sin \left (f x +e \right )+46\right )}{21 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(142\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\frac {2 \, {\left (15 \, B a^{2} \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - {\left (77 \, A + 85 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (161 \, A + 145 \, B\right )} a^{2} \cos \left (f x + e\right ) - 32 \, {\left (7 \, A + 5 \, B\right )} a^{2} + {\left (15 \, B a^{2} \cos \left (f x + e\right )^{3} - 3 \, {\left (7 \, A + 15 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (49 \, A + 65 \, B\right )} a^{2} \cos \left (f x + e\right ) + 32 \, {\left (7 \, A + 5 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

2/105*(15*B*a^2*cos(f*x + e)^4 + 3*(7*A + 20*B)*a^2*cos(f*x + e)^3 - (77*A + 85*B)*a^2*cos(f*x + e)^2 - 2*(161
*A + 145*B)*a^2*cos(f*x + e) - 32*(7*A + 5*B)*a^2 + (15*B*a^2*cos(f*x + e)^3 - 3*(7*A + 15*B)*a^2*cos(f*x + e)
^2 - 2*(49*A + 65*B)*a^2*cos(f*x + e) + 32*(7*A + 5*B)*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x
+ e) + f*sin(f*x + e) + f)

Sympy [F]

\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(5/2)*(A + B*sin(e + f*x)), x)

Maxima [F]

\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.46 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (15 \, B a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 525 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (10 \, A a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 11 \, B a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \]

[In]

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

[Out]

1/420*sqrt(2)*(15*B*a^2*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*a^2*sgn(
cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a^2*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*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 11*B*a^2*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*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*B*a^2*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(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2), x)