3.180 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=271 \[ -\frac{2 c^3 (3 A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a \sin (e+f x)+a}}-\frac{c^2 (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a \sin (e+f x)+a}}-\frac{4 c^4 (3 A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

(-4*(3*A - 5*B)*c^4*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]
) - (2*(3*A - 5*B)*c^3*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c^
2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(5/2))/(6*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(
2*f*(a + a*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.577265, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2972, 2740, 2737, 2667, 31} \[ -\frac{2 c^3 (3 A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a \sin (e+f x)+a}}-\frac{c^2 (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a \sin (e+f x)+a}}-\frac{4 c^4 (3 A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-4*(3*A - 5*B)*c^4*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]
) - (2*(3*A - 5*B)*c^3*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c^
2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(5/2))/(6*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(
2*f*(a + a*Sin[e + f*x])^(3/2))

Rule 2972

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), 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, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 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 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{(3 A-5 B) \int \frac{(c-c \sin (e+f x))^{7/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{((3 A-5 B) c) \int \frac{(c-c \sin (e+f x))^{5/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left (2 (3 A-5 B) c^2\right ) \int \frac{(c-c \sin (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{2 (3 A-5 B) c^3 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 (3 A-5 B) c^3\right ) \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{2 (3 A-5 B) c^3 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 (3 A-5 B) c^4 \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 (3 A-5 B) c^3 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 (3 A-5 B) c^4 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{4 (3 A-5 B) c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}-\frac{2 (3 A-5 B) c^3 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.51203, size = 271, normalized size = 1. \[ -\frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 (27 A-59 B) \cos (2 (e+f x))-117 A \sin (e+f x)-3 A \sin (3 (e+f x))+576 A \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+576 A \sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+132 A+279 B \sin (e+f x)+13 B \sin (3 (e+f x))+B \cos (4 (e+f x))-960 B \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-960 B \sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-45 B\right )}{24 f (a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(132*A - 45*B + 2*(27*A - 59*B)*Cos[2*(e
+ f*x)] + B*Cos[4*(e + f*x)] + 576*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 960*B*Log[Cos[(e + f*x)/2] + S
in[(e + f*x)/2]] - 117*A*Sin[e + f*x] + 279*B*Sin[e + f*x] + 576*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Si
n[e + f*x] - 960*B*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 3*A*Sin[3*(e + f*x)] + 13*B*Sin[3*(
e + f*x)]))/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B]  time = 0.288, size = 938, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x)

[Out]

1/6/f*(-102*A+166*B-102*A*sin(f*x+e)+288*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-480*B*ln((1-cos(f*x+e)+sin
(f*x+e))/sin(f*x+e))+99*A*cos(f*x+e)^2+24*A*cos(f*x+e)^2*sin(f*x+e)+72*A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))-120*B
*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+2*B*sin(f*x+e)*cos(f*x+e)^4+3*A*cos(f*x+e)^3*sin(f*x+e)-48*B*cos(f*x+e)^2*s
in(f*x+e)+27*A*cos(f*x+e)+72*A*cos(f*x+e)*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-13*B*cos(f*x+e)^3*sin(f*x+e)-27*A*co
s(f*x+e)^3+61*B*cos(f*x+e)^3-59*B*cos(f*x+e)-120*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)-144*A*sin(f*x+e)
*ln(2/(cos(f*x+e)+1))-107*B*sin(f*x+e)*cos(f*x+e)-120*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+240*B*sin(f*x+e)*ln(2/
(cos(f*x+e)+1))+75*A*sin(f*x+e)*cos(f*x+e)+72*A*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-2*B*cos(f*x+e)^5+240*B*ln((1
-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-144*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-1
44*A*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+240*B*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x
+e))+288*A*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-480*B*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/s
in(f*x+e))+3*A*cos(f*x+e)^4+240*B*cos(f*x+e)*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-144*A*cos(f*x
+e)*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-11*B*cos(f*x+e)^4-155*B*cos(f*x+e)^2-144*A*ln(2/(cos(f
*x+e)+1))+240*B*ln(2/(cos(f*x+e)+1))+166*B*sin(f*x+e))*(-c*(-1+sin(f*x+e)))^(7/2)/(cos(f*x+e)^4-sin(f*x+e)*cos
(f*x+e)^3+3*cos(f*x+e)^3+4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2+4*sin(f*x+e)*cos(f*x+e)-4*cos(f*x+e)-8*sin(f
*x+e)+8)/(a*(1+sin(f*x+e)))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B c^{3} \cos \left (f x + e\right )^{4} +{\left (3 \, A - 5 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 4 \,{\left (A - B\right )} c^{3} -{\left ({\left (A - 3 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 4 \,{\left (A - B\right )} 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}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((B*c^3*cos(f*x + e)^4 + (3*A - 5*B)*c^3*cos(f*x + e)^2 - 4*(A - B)*c^3 - ((A - 3*B)*c^3*cos(f*x + e)^
2 - 4*(A - B)*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^
2*sin(f*x + e) - 2*a^2), x)

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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+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(3/2), x)