3.104 \(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=225 \[ \frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a^3 c (3 A+11 B) \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/4*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^3*(3*A+11*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^
(7/2)-5/24*a^3*(3*A+11*B)*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^(3/2)+5/4*a^3*(3*A+11*B)*arctanh(1/2*cos(f*x+e)*c^
(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)-5/4*a^3*(3*A+11*B)*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(
1/2)

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Rubi [A]  time = 0.55, antiderivative size = 225, 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} \[ \frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}-\frac {a^3 c (3 A+11 B) \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}} \]

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])^(5/2),x]

[Out]

(5*a^3*(3*A + 11*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f)
+ (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(4*f*(c - c*Sin[e + f*x])^(11/2)) - (a^3*(3*A + 11*B)*c*Cos[e + f*x]^5)/(8*
f*(c - c*Sin[e + f*x])^(7/2)) - (5*a^3*(3*A + 11*B)*Cos[e + f*x]^3)/(24*c*f*(c - c*Sin[e + f*x])^(3/2)) - (5*a
^3*(3*A + 11*B)*Cos[e + f*x])/(4*c^2*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))^{5/2}} \, 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))^{11/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {1}{8} \left (a^3 (3 A+11 B) c^2\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}+\frac {1}{16} \left (5 a^3 (3 A+11 B)\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}+\frac {\left (5 a^3 (3 A+11 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 c}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (5 a^3 (3 A+11 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{4 c^2}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (5 a^3 (3 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 )}{2 c^2 f}\\ &=\frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 2.21, size = 434, normalized size = 1.93 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-6 (2 A+11 B) \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-6 (2 A+11 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-3 (9 A+17 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-6 (9 A+17 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(-15-15 i) \sqrt [4]{-1} (3 A+11 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+2 B \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-2 B \sin \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{6 f (c-c \sin (e+f x))^{5/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

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])^(5/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]) - 3*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1/4)*(3*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])^4 - 6*(2*A + 11*B)*Cos[(e + f
*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 2*B*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]
)^4 + 24*(A + B)*Sin[(e + f*x)/2] - 6*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] -
6*(2*A + 11*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 2*B*(Cos[(e + f*x)/2] - Sin[(e + f*x
)/2])^4*Sin[(3*(e + f*x))/2]))/(6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(5/2))

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fricas [B]  time = 0.46, size = 505, normalized size = 2.24 \[ \frac {15 \, \sqrt {2} {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, B\right )} a^{3} - {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, 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 (4 \, B a^{3} \cos \left (f x + e\right )^{4} - 4 \, {\left (3 \, A + 14 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (13 \, A + 37 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (13 \, A + 53 \, B\right )} a^{3} \cos \left (f x + e\right ) - 12 \, {\left (A + B\right )} a^{3} - {\left (4 \, B a^{3} \cos \left (f x + e\right )^{3} + 12 \, {\left (A + 5 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (17 \, A + 57 \, B\right )} a^{3} \cos \left (f x + e\right ) + 12 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{24 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/24*(15*sqrt(2)*((3*A + 11*B)*a^3*cos(f*x + e)^3 + 3*(3*A + 11*B)*a^3*cos(f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos
(f*x + e) - 4*(3*A + 11*B)*a^3 - ((3*A + 11*B)*a^3*cos(f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos(f*x + e) - 4*(3*A +
11*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*(4*B*a^3*cos(f*x + e)^4 - 4*(3*A + 14*B)*a^3*cos(f*x +
e)^3 + 3*(13*A + 37*B)*a^3*cos(f*x + e)^2 + 3*(13*A + 53*B)*a^3*cos(f*x + e) - 12*(A + B)*a^3 - (4*B*a^3*cos(
f*x + e)^3 + 12*(A + 5*B)*a^3*cos(f*x + e)^2 + 3*(17*A + 57*B)*a^3*cos(f*x + e) + 12*(A + B)*a^3)*sin(f*x + e)
)*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f -
(c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

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

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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 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 si
gn: (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/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*(2/sqrt(c*tan((f*x+exp(1))/2)^2+c)/(
c*tan((f*x+exp(1))/2)^2+c)*(tan((f*x+exp(1))/2)*(tan((f*x+exp(1))/2)*(-1/144*tan((f*x+exp(1))/2)*(-72*A*a^3*c^
3*sign(tan((f*x+exp(1))/2)-1)-384*B*a^3*c^3*sign(tan((f*x+exp(1))/2)-1))/c^4-1/144*(-72*A*a^3*c^3*sign(tan((f*
x+exp(1))/2)-1)-432*B*a^3*c^3*sign(tan((f*x+exp(1))/2)-1))/c^4)-1/144*(-72*A*a^3*c^3*sign(tan((f*x+exp(1))/2)-
1)-432*B*a^3*c^3*sign(tan((f*x+exp(1))/2)-1))/c^4)-1/144*(-72*A*a^3*c^3*sign(tan((f*x+exp(1))/2)-1)-384*B*a^3*
c^3*sign(tan((f*x+exp(1))/2)-1))/c^4)+2*(1/4*(7*A*a^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)
^2+c))^7+31*B*a^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7+81*A*a^3*sqrt(c)*(-sqrt(c)*
tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6+185*B*a^3*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))^6+53*A*a^3*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+125*B*
a^3*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-65*A*a^3*sqrt(c)*c*(-sqrt(c)*tan((f*x+e
xp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4-169*B*a^3*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x
+exp(1))/2)^2+c))^4+13*A*a^3*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+21*B*a^3*c^2
*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-33*A*a^3*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+s
qrt(c*tan((f*x+exp(1))/2)^2+c))+19*A*a^3*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^
2+c))^2-73*B*a^3*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+75*B*a^3*sqrt(c)*c^2*(-sqr
t(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+5*A*a^3*sqrt(c)*c^3+13*B*a^3*sqrt(c)*c^3)/c^2/(-(-
sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))+c)^4/sign(tan((f*x+exp(1))/2)-1)+1/4*(15*A*a^3+55*B*a^3)*atan((-sqrt(c)*tan((f*x+exp
(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/c^2/sqrt(-c)/sign(tan((f*x+exp(1))/
2)-1)))

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maple [B]  time = 1.99, size = 434, normalized size = 1.93 \[ -\frac {a^{3} \left (\sin \left (f x +e \right ) \left (-90 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+48 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-330 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+240 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+16 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}\right )+\left (-45 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+24 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-165 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+120 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+8 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+90 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-132 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+54 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+330 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-420 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+86 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{12 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^(5/2),x)

[Out]

-1/12/c^(9/2)*a^3*(sin(f*x+e)*(-90*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+48*A*(c+c
*sin(f*x+e))^(1/2)*c^(3/2)-330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+240*B*(c+c*si
n(f*x+e))^(1/2)*c^(3/2)+16*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2))+(-45*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)
*2^(1/2)/c^(1/2))*c^2+24*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-165*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(
1/2)/c^(1/2))*c^2+120*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2)+8*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2))*cos(f*x+e)^2+90*A*2
^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2-132*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)+54*A*(c+c*
sin(f*x+e))^(3/2)*c^(1/2)+330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2-420*B*(c+c*sin
(f*x+e))^(1/2)*c^(3/2)+86*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/cos(f*x+e)
/(c-c*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

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))^(5/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(5/2), x)

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

[Out]

Timed out

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