3.149 \(\int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=198 \[ -\frac{2 a^2 (7 A-B) \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{105 f}-\frac{a^3 (7 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{a (7 A-B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \]

[Out]

-(a^3*(7*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*(7*A - B)*C
os[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(105*f) - (a*(7*A - B)*Cos[e + f*x]*(a + a*Si
n[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2))/(42*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e
+ f*x])^(7/2))/(7*f)

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Rubi [A]  time = 0.476191, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ -\frac{2 a^2 (7 A-B) \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{105 f}-\frac{a^3 (7 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{a (7 A-B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a^3*(7*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*(7*A - B)*C
os[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(105*f) - (a*(7*A - B)*Cos[e + f*x]*(a + a*Si
n[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2))/(42*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e
+ f*x])^(7/2))/(7*f)

Rule 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2738

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] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac{1}{7} (7 A-B) \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx\\ &=-\frac{a (7 A-B) \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac{1}{21} (2 a (7 A-B)) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx\\ &=-\frac{2 a^2 (7 A-B) \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{105 f}-\frac{a (7 A-B) \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac{1}{105} \left (4 a^2 (7 A-B)\right ) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx\\ &=-\frac{a^3 (7 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{105 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 (7 A-B) \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{105 f}-\frac{a (7 A-B) \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\\ \end{align*}

Mathematica [A]  time = 2.85785, size = 223, normalized size = 1.13 \[ -\frac{c^3 (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)} (525 (A-B) \cos (2 (e+f x))+210 (A-B) \cos (4 (e+f x))+4200 A \sin (e+f x)+700 A \sin (3 (e+f x))+84 A \sin (5 (e+f x))+35 A \cos (6 (e+f x))-525 B \sin (e+f x)+35 B \sin (3 (e+f x))+63 B \sin (5 (e+f x))+15 B \sin (7 (e+f x))-35 B \cos (6 (e+f x)))}{6720 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(525*(A - B)*Cos[2*(e + f*x)
] + 210*(A - B)*Cos[4*(e + f*x)] + 35*A*Cos[6*(e + f*x)] - 35*B*Cos[6*(e + f*x)] + 4200*A*Sin[e + f*x] - 525*B
*Sin[e + f*x] + 700*A*Sin[3*(e + f*x)] + 35*B*Sin[3*(e + f*x)] + 84*A*Sin[5*(e + f*x)] + 63*B*Sin[5*(e + f*x)]
+ 15*B*Sin[7*(e + f*x)]))/(6720*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2
])^5)

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Maple [A]  time = 0.368, size = 203, normalized size = 1. \begin{align*}{\frac{ \left ( -30\,B \left ( \cos \left ( fx+e \right ) \right ) ^{6}+35\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sin \left ( fx+e \right ) -35\,B\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-42\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+35\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -35\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -56\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+35\,A\sin \left ( fx+e \right ) -35\,B\sin \left ( fx+e \right ) -112\,A+16\,B \right ) \sin \left ( fx+e \right ) }{210\,f \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}}} \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))*(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/210/f*(-30*B*cos(f*x+e)^6+35*A*cos(f*x+e)^4*sin(f*x+e)-35*B*sin(f*x+e)*cos(f*x+e)^4-42*A*cos(f*x+e)^4+6*B*co
s(f*x+e)^4+35*A*cos(f*x+e)^2*sin(f*x+e)-35*B*cos(f*x+e)^2*sin(f*x+e)-56*A*cos(f*x+e)^2+8*B*cos(f*x+e)^2+35*A*s
in(f*x+e)-35*B*sin(f*x+e)-112*A+16*B)*(-c*(-1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/(-1+sin(f
*x+e))/cos(f*x+e)^5

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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}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2), x)

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Fricas [A]  time = 2.34748, size = 371, normalized size = 1.87 \begin{align*} \frac{{\left (35 \,{\left (A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{6} - 35 \,{\left (A - B\right )} a^{2} c^{3} + 2 \,{\left (15 \, B a^{2} c^{3} \cos \left (f x + e\right )^{6} + 3 \,{\left (7 \, A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{4} + 4 \,{\left (7 \, A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{2} + 8 \,{\left (7 \, A - B\right )} a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{210 \, f \cos \left (f x + e\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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/210*(35*(A - B)*a^2*c^3*cos(f*x + e)^6 - 35*(A - B)*a^2*c^3 + 2*(15*B*a^2*c^3*cos(f*x + e)^6 + 3*(7*A - B)*a
^2*c^3*cos(f*x + e)^4 + 4*(7*A - B)*a^2*c^3*cos(f*x + e)^2 + 8*(7*A - B)*a^2*c^3)*sin(f*x + e))*sqrt(a*sin(f*x
+ e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

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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))*(c-c*sin(f*x+e))**(7/2),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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

Timed out