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

Optimal. Leaf size=217 \[ -\frac {5 a^3 (A+13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}+\frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}+\frac {5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt {c-c \sin (e+f x)}}-\frac {a^3 c (A+13 B) \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/6*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(13/2)-1/24*a^3*(A+13*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(
9/2)+5/48*a^3*(A+13*B)*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^(5/2)-5/16*a^3*(A+13*B)*arctanh(1/2*cos(f*x+e)*c^(1/2
)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(7/2)/f*2^(1/2)+5/16*a^3*(A+13*B)*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.55, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2967, 2859, 2680, 2679, 2649, 206} \[ \frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}+\frac {5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt {c-c \sin (e+f x)}}-\frac {5 a^3 (A+13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}-\frac {a^3 c (A+13 B) \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^3*(A + 13*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(7/2)*f) +
(a^3*(A + B)*c^3*Cos[e + f*x]^7)/(6*f*(c - c*Sin[e + f*x])^(13/2)) - (a^3*(A + 13*B)*c*Cos[e + f*x]^5)/(24*f*
(c - c*Sin[e + f*x])^(9/2)) + (5*a^3*(A + 13*B)*Cos[e + f*x]^3)/(48*c*f*(c - c*Sin[e + f*x])^(5/2)) + (5*a^3*(
A + 13*B)*Cos[e + f*x])/(16*c^3*f*Sqrt[c - c*Sin[e + f*x]])

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])

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 2679

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

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 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 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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {1}{12} \left (a^3 (A+13 B) c^2\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {1}{48} \left (5 a^3 (A+13 B)\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}-\frac {\left (5 a^3 (A+13 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{32 c^2}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac {5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (5 a^3 (A+13 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 c^3}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac {5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (5 a^3 (A+13 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 )}{8 c^3 f}\\ &=-\frac {5 a^3 (A+13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac {5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 3.09, size = 422, normalized size = 1.94 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (64 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )+3 (11 A+47 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+6 (11 A+47 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-4 (13 A+25 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-8 (13 A+25 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+32 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} (A+13 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+48 B \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+48 B \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6\right )}{24 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(13*A + 25*B)
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 3*(11*A + 47*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + (15 + 15*
I)*(-1)^(1/4)*(A + 13*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 + 48*B*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6 + 64*(A + B)*Sin[(e + f*x)/2] - 8*(13
*A + 25*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 6*(11*A + 47*B)*(Cos[(e + f*x)/2] - Sin[
(e + f*x)/2])^4*Sin[(e + f*x)/2] + 48*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2])*(1 + Sin[e +
f*x])^3)/(24*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(7/2))

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 554, normalized size = 2.55 \[ \frac {15 \, \sqrt {2} {\left ({\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 3 \, {\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 8 \, {\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right ) + 8 \, {\left (A + 13 \, B\right )} a^{3} + {\left ({\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 4 \, {\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (A + 13 \, B\right )} a^{3} \cos \left (f x + e\right ) - 8 \, {\left (A + 13 \, B\right )} a^{3}\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 (48 \, B a^{3} \cos \left (f x + e\right )^{4} + 3 \, {\left (11 \, A + 95 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + {\left (19 \, A - 137 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (23 \, A + 203 \, B\right )} a^{3} \cos \left (f x + e\right ) - 32 \, {\left (A + B\right )} a^{3} - {\left (48 \, B a^{3} \cos \left (f x + e\right )^{3} - 3 \, {\left (11 \, A + 79 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, A + 187 \, B\right )} a^{3} \cos \left (f x + e\right ) + 32 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, {\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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(15*sqrt(2)*((A + 13*B)*a^3*cos(f*x + e)^4 - 3*(A + 13*B)*a^3*cos(f*x + e)^3 - 8*(A + 13*B)*a^3*cos(f*x +
e)^2 + 4*(A + 13*B)*a^3*cos(f*x + e) + 8*(A + 13*B)*a^3 + ((A + 13*B)*a^3*cos(f*x + e)^3 + 4*(A + 13*B)*a^3*c
os(f*x + e)^2 - 4*(A + 13*B)*a^3*cos(f*x + e) - 8*(A + 13*B)*a^3)*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*
(48*B*a^3*cos(f*x + e)^4 + 3*(11*A + 95*B)*a^3*cos(f*x + e)^3 + (19*A - 137*B)*a^3*cos(f*x + e)^2 - 2*(23*A +
203*B)*a^3*cos(f*x + e) - 32*(A + B)*a^3 - (48*B*a^3*cos(f*x + e)^3 - 3*(11*A + 79*B)*a^3*cos(f*x + e)^2 - 2*(
7*A + 187*B)*a^3*cos(f*x + e) + 32*(A + B)*a^3)*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))

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by intervals (
correct if the argument is real):Check [abs(sin((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (8*pi/t_nos
tep/2)>(-8*pi/t_nostep/2)Discontinuities at zeroes of sin((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable t
o check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Un
able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to
check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_no
step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(t_nostep-1)]Evaluation time: 2.23Not inver
tible Error: Bad Argument Value

________________________________________________________________________________________

maple [B]  time = 1.98, size = 524, normalized size = 2.41 \[ \frac {a^{3} \left (15 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{3}+195 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{3}-45 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{3}-96 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}} \left (\sin ^{3}\left (f x +e \right )\right )-585 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{3}+66 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {c}+45 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{3}+282 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {c}+288 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}} \left (\sin ^{2}\left (f x +e \right )\right )+585 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{3}-160 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-15 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-928 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-288 B \,c^{\frac {5}{2}} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sin \left (f x +e \right )-195 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+120 A \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}}+888 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{48 c^{\frac {13}{2}} \left (\sin \left (f x +e \right )-1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48/c^(13/2)*a^3*(15*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^3+195*B*2
^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^3-45*A*2^(1/2)*arctanh(1/2*(c*(1+s
in(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^3-96*B*(c*(1+sin(f*x+e)))^(1/2)*c^(5/2)*sin(f*x+e)^3-585*B*2
^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^3+66*A*(c*(1+sin(f*x+e)))^(5/2)*c^
(1/2)+45*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^3+282*B*(c*(1+sin(f*x+e)
))^(5/2)*c^(1/2)+288*B*(c*(1+sin(f*x+e)))^(1/2)*c^(5/2)*sin(f*x+e)^2+585*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e
)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^3-160*A*(c*(1+sin(f*x+e)))^(3/2)*c^(3/2)-15*A*2^(1/2)*arctanh(1/2*(c*(
1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3-928*B*(c*(1+sin(f*x+e)))^(3/2)*c^(3/2)-288*B*c^(5/2)*(c*(1+sin(f*x+e
)))^(1/2)*sin(f*x+e)-195*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3+120*A*(c*(1+sin(f
*x+e)))^(1/2)*c^(5/2)+888*B*(c*(1+sin(f*x+e)))^(1/2)*c^(5/2))*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)^2/cos(f*
x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________