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

Optimal. Leaf size=263 \[ -\frac{3 a^3 (A+3 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{3 a^2 (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{6 a^4 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.595239, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2972, 2739, 2740, 2737, 2667, 31} \[ -\frac{3 a^3 (A+3 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{3 a^2 (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{6 a^4 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]

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])^(5/2),x]

[Out]

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

Mathematica [A]  time = 2.59541, size = 251, normalized size = 0.95 \[ \frac{(a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-4 (A+6 B) \sin (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-16 (3 A+5 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2-48 (A+3 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+16 (A+B)+B \cos (2 (e+f x)) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{4 f (c-c \sin (e+f x))^{5/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2)*(16*(A + B) - 16*(3*A + 5*B)*(Cos[(e + f*x
)/2] - Sin[(e + f*x)/2])^2 + B*Cos[2*(e + f*x)]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 48*(A + 3*B)*Log[Cos
[(e + f*x)/2] - Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 4*(A + 6*B)*(Cos[(e + f*x)/2] - Si
n[(e + f*x)/2])^4*Sin[e + f*x]))/(4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.28, size = 1189, normalized size = 4.5 \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))^(5/2),x)

[Out]

-1/2/f*(32*A+100*B-32*A*sin(f*x+e)-34*A*cos(f*x+e)^2+22*A*cos(f*x+e)^2*sin(f*x+e)-24*A*cos(f*x+e)*ln(2/(cos(f*
x+e)+1))-108*B*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+B*sin(f*x+e)*cos(f*x+e)^4-2*A*cos(f*x+e)^3*sin(f*x+e)+63*B*co
s(f*x+e)^2*sin(f*x+e)-20*A*cos(f*x+e)-48*A*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+24
*A*cos(f*x+e)*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-10*B*cos(f*x+e)^3*sin(f*x+e)+48*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)+
sin(f*x+e))/sin(f*x+e))+216*B*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+20*A*cos(f*x+e)^3+53*B*c
os(f*x+e)^3-54*B*cos(f*x+e)+72*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)-144*B*ln(-(-1+cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)-48*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+96*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)+s
in(f*x+e))/sin(f*x+e))+46*B*sin(f*x+e)*cos(f*x+e)-72*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+144*B*cos(f*x+e)*ln(-(-
1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-144*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+288*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)
+sin(f*x+e))/sin(f*x+e))+12*A*sin(f*x+e)*cos(f*x+e)+72*A*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))-36*A*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+B*cos(f*x+e)^5-24*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f
*x+e)^3+12*A*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^3-24*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*si
n(f*x+e)+12*A*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^2*sin(f*x+e)+2*A*cos(f*x+e)^4+9*B*cos(f*x+e)^4-109*B*cos(f*x+e)^
2+48*A*ln(2/(cos(f*x+e)+1))-96*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+144*B*ln(2/(cos(f*x+e)+1))-288*B*l
n(-(-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))+36*B*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-72*B*cos(f*x+e)^3
*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-100*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)))^(5/2)

________________________________________________________________________________________

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{5}{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))^(5/2),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)^(5/2), x)

________________________________________________________________________________________

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}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\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))^(5/2),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)/(3*c^3*cos(f*x + e)^2 - 4
*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)

________________________________________________________________________________________

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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{5}{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))^(5/2),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)^(5/2), x)