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

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

[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)

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Rubi [A]  time = 0.112403, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2751, 2647, 2646} \[ -\frac{64 a^3 (7 A+5 B) \cos (e+f x)}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{16 a^2 (7 A+5 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{105 f}-\frac{2 a (7 A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac{2 B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f} \]

Antiderivative was successfully verified.

[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 2751

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*Sin
[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 2647

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] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx &=-\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]  time = 1.5225, size = 119, normalized size = 0.86 \[ -\frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) ((392 A+505 B) \sin (e+f x)-6 (7 A+20 B) \cos (2 (e+f x))+1246 A-15 B \sin (3 (e+f x))+1040 B)}{210 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)]))/(210*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]))

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Maple [A]  time = 0.897, size = 99, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ){a}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( -15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( 98\,A+130\,B \right ) \sin \left ( fx+e \right ) + \left ( -21\,A-60\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+322\,A+290\,B \right ) }{105\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.61816, size = 483, normalized size = 3.5 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out