3.169 \(\int \frac{(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=247 \[ -\frac{a^3 B \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^4 B \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) - (a*B*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(5/2))/(3*c*f*(c - c*Sin[e + f*x])^(7/2)) + (a^2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*c
^2*f*(c - c*Sin[e + f*x])^(5/2)) - (a^3*B*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c^3*f*(c - c*Sin[e + f*x])^(
3/2)) - (a^4*B*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(c^4*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.598736, antiderivative size = 247, 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, 2739, 2737, 2667, 31} \[ -\frac{a^3 B \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^4 B \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) - (a*B*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(5/2))/(3*c*f*(c - c*Sin[e + f*x])^(7/2)) + (a^2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*c
^2*f*(c - c*Sin[e + f*x])^(5/2)) - (a^3*B*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c^3*f*(c - c*Sin[e + f*x])^(
3/2)) - (a^4*B*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(c^4*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

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 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[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+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{B \int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{(a B) \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c^2}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{a^2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{\left (a^2 B\right ) \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^3}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{a^2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^3 B\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx}{c^4}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{a^2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^4 B \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^3 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{a^2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^4 B \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^4 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 f (c-c \sin (e+f x))^{9/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}+\frac{a^2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}-\frac{a^4 B \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.74836, size = 238, normalized size = 0.96 \[ \frac{(a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-3 (A+7 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+9 (A+3 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-4 (3 A+5 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+6 (A+B)-6 B \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 f (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

((6*(A + B) - 4*(3*A + 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + 9*(A + 3*B)*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])^4 - 3*(A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6 - 6*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2
]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2)
)/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(9/2))

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Maple [B]  time = 0.323, size = 1019, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/3/f*(6*A-34*B-6*A*sin(f*x+e)-9*A*cos(f*x+e)^2+3*A*cos(f*x+e)^2*sin(f*x+e)+60*B*cos(f*x+e)^2*ln(2/(cos(f*x+e)
+1))+8*B*sin(f*x+e)*cos(f*x+e)^4-3*A*cos(f*x+e)^3*sin(f*x+e)-39*B*cos(f*x+e)^2*sin(f*x+e)+3*B*cos(f*x+e)^4*sin
(f*x+e)*ln(2/(cos(f*x+e)+1))+11*B*cos(f*x+e)^3*sin(f*x+e)-120*B*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))-28*B*cos(f*x+e)^3+20*B*cos(f*x+e)-24*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)+48*B*ln(-(-1+cos(f
*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)-14*B*sin(f*x+e)*cos(f*x+e)+24*B*cos(f*x+e)*ln(2/(cos(f*x+e
)+1))-48*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+48*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-96*B*sin
(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+6*A*sin(f*x+e)*cos(f*x+e)-6*B*cos(f*x+e)^4*sin(f*x+e)*ln(-(
-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-15*B*cos(f*x+e)^4*ln(2/(cos(f*x+e)+1))+30*B*cos(f*x+e)^4*ln(-(-1+cos(f*x
+e)+sin(f*x+e))/sin(f*x+e))+3*B*cos(f*x+e)^5*ln(2/(cos(f*x+e)+1))+8*B*cos(f*x+e)^5-24*B*cos(f*x+e)^3*sin(f*x+e
)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+12*B*cos(f*x+e)^3*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-6*B*cos(f*x+e)^
5*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*A*cos(f*x+e)^4-19*B*cos(f*x+e)^4+53*B*cos(f*x+e)^2-48*B*ln(2/(c
os(f*x+e)+1))+96*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+72*B*cos(f*x+e)^2*sin(f*x+e)*ln(-(-1+cos(f*x+e)+
sin(f*x+e))/sin(f*x+e))-36*B*cos(f*x+e)^2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-24*B*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1
))+48*B*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+34*B*sin(f*x+e))*(a*(1+sin(f*x+e)))^(7/2)/(sin
(f*x+e)*cos(f*x+e)^3+cos(f*x+e)^4-4*cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^3-4*sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e
)^2+8*sin(f*x+e)-4*cos(f*x+e)+8)/(-c*(-1+sin(f*x+e)))^(9/2)

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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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B a^{3} \cos \left (f x + e\right )^{4} -{\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 4 \,{\left (A + B\right )} a^{3} -{\left ({\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 4 \,{\left (A + B\right )} a^{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}}{5 \, c^{5} \cos \left (f x + e\right )^{4} - 20 \, c^{5} \cos \left (f x + e\right )^{2} + 16 \, c^{5} -{\left (c^{5} \cos \left (f x + e\right )^{4} - 12 \, c^{5} \cos \left (f x + e\right )^{2} + 16 \, c^{5}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

integral((B*a^3*cos(f*x + e)^4 - (3*A + 5*B)*a^3*cos(f*x + e)^2 + 4*(A + B)*a^3 - ((A + 3*B)*a^3*cos(f*x + e)^
2 - 4*(A + B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^5*cos(f*x + e)^4 - 20
*c^5*cos(f*x + e)^2 + 16*c^5 - (c^5*cos(f*x + e)^4 - 12*c^5*cos(f*x + e)^2 + 16*c^5)*sin(f*x + e)), 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+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2),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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(9/2), x)