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

Optimal. Leaf size=175 \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}-\frac{a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 (A-11 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

(a^2*(A - 11*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(6*f*(c - c*Sin[e + f*x])^(11/2)) + (a^2*(A - 11*B)*Cos[e + f*x]^3)/(24*f*(c -
c*Sin[e + f*x])^(7/2)) - (a^2*(A - 11*B)*Cos[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.491015, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {2967, 2859, 2680, 2649, 206} \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}-\frac{a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 (A-11 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(A - 11*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(6*f*(c - c*Sin[e + f*x])^(11/2)) + (a^2*(A - 11*B)*Cos[e + f*x]^3)/(24*f*(c -
c*Sin[e + f*x])^(7/2)) - (a^2*(A - 11*B)*Cos[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(3/2))

Rule 2967

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

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac{1}{12} \left (a^2 (A-11 B) c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac{\left (a^2 (A-11 B)\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{16 c}\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^2 (A-11 B)\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{32 c^3}\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^2 (A-11 B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{16 c^3 f}\\ &=\frac{a^2 (A-11 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac{a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.83863, size = 342, normalized size = 1.95 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )+3 (A+21 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+6 (A+21 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-4 (7 A+19 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-8 (7 A+19 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+32 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(-3-3 i) \sqrt [4]{-1} (A-11 B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6\right )}{48 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(7*A + 19*B)*
(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 3*(A + 21*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - (3 + 3*I)*(-1
)^(1/4)*(A - 11*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
^6 + 64*(A + B)*Sin[(e + f*x)/2] - 8*(7*A + 19*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 6
*(A + 21*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(48*f*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(7/2))

________________________________________________________________________________________

Maple [B]  time = 1.628, size = 354, normalized size = 2. \begin{align*} -{\frac{{a}^{2}}{96\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( -3\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{3} \left ( A-11\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+12\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{3} \left ( A-11\,B \right ) \sin \left ( fx+e \right ) +9\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{3} \left ( A-11\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+24\,A\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{5/2}-32\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{3/2}-6\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}\sqrt{c}-264\,B\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{5/2}+352\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{3/2}-126\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}\sqrt{c}-12\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{3}+132\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{3} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{13}{2}}}{\frac{1}{\sqrt{c-c\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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x)

[Out]

-1/96*a^2*(-3*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*(A-11*B)*sin(f*x+e)*cos(f*x+e)^2
+12*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*(A-11*B)*sin(f*x+e)+9*arctanh(1/2*(c+c*sin
(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*(A-11*B)*cos(f*x+e)^2+24*A*(c+c*sin(f*x+e))^(1/2)*c^(5/2)-32*A*(c+
c*sin(f*x+e))^(3/2)*c^(3/2)-6*A*(c+c*sin(f*x+e))^(5/2)*c^(1/2)-264*B*(c+c*sin(f*x+e))^(1/2)*c^(5/2)+352*B*(c+c
*sin(f*x+e))^(3/2)*c^(3/2)-126*B*(c+c*sin(f*x+e))^(5/2)*c^(1/2)-12*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2
)*2^(1/2)/c^(1/2))*c^3+132*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^3)*(c*(1+sin(f*x+e)
))^(1/2)/c^(13/2)/(-1+sin(f*x+e))^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

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 )}^{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))^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)^2/(-c*sin(f*x + e) + c)^(7/2), x)

________________________________________________________________________________________

Fricas [B]  time = 1.59873, size = 1353, normalized size = 7.73 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} - 3 \,{\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - 8 \,{\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 4 \,{\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right ) + 8 \,{\left (A - 11 \, B\right )} a^{2} +{\left ({\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 4 \,{\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 4 \,{\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right ) - 8 \,{\left (A - 11 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \,{\left (3 \,{\left (A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} +{\left (25 \, A + 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (5 \, A + 41 \, B\right )} a^{2} \cos \left (f x + e\right ) - 32 \,{\left (A + B\right )} a^{2} +{\left (3 \,{\left (A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (11 \, A - 25 \, B\right )} a^{2} \cos \left (f x + e\right ) - 32 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{192 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(3*sqrt(2)*((A - 11*B)*a^2*cos(f*x + e)^4 - 3*(A - 11*B)*a^2*cos(f*x + e)^3 - 8*(A - 11*B)*a^2*cos(f*x
+ e)^2 + 4*(A - 11*B)*a^2*cos(f*x + e) + 8*(A - 11*B)*a^2 + ((A - 11*B)*a^2*cos(f*x + e)^3 + 4*(A - 11*B)*a^2*
cos(f*x + e)^2 - 4*(A - 11*B)*a^2*cos(f*x + e) - 8*(A - 11*B)*a^2)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^
2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(
f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4
*(3*(A + 21*B)*a^2*cos(f*x + e)^3 + (25*A + 13*B)*a^2*cos(f*x + e)^2 - 2*(5*A + 41*B)*a^2*cos(f*x + e) - 32*(A
+ B)*a^2 + (3*(A + 21*B)*a^2*cos(f*x + e)^2 - 2*(11*A - 25*B)*a^2*cos(f*x + e) - 32*(A + B)*a^2)*sin(f*x + e)
)*sqrt(-c*sin(f*x + e) + c))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f
*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*f)*sin
(f*x + e))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2