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

Optimal. Leaf size=176 \[ -\frac {\sqrt {2} a^2 (3 A+7 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 (3 A+7 B) \cos ^3(e+f x)}{6 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 (3 A+7 B) \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 176, normalized size of antiderivative = 1.00, 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, 2679, 2649, 206} \[ \frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {\sqrt {2} a^2 (3 A+7 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 (3 A+7 B) \cos ^3(e+f x)}{6 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 (3 A+7 B) \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \]

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])^(3/2),x]

[Out]

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

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Mathematica [C]  time = 0.90, size = 355, normalized size = 2.02 \[ \frac {a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )+3 (2 A+7 B) \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+3 (2 A+7 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+6 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(6+6 i) \sqrt [4]{-1} (3 A+7 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-B \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+B \sin \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )}{3 f (c-c \sin (e+f x))^{3/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4} \]

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])^(3/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*(6*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2
]) + (6 + 6*I)*(-1)^(1/4)*(3*A + 7*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])^2 + 3*(2*A + 7*B)*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 - B*Cos[(3*(e +
f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + 12*(A + B)*Sin[(e + f*x)/2] + 3*(2*A + 7*B)*(Cos[(e + f*x)
/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(3*(e + f*x))/2]))/
(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(3/2))

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fricas [B]  time = 0.45, size = 385, normalized size = 2.19 \[ \frac {\frac {3 \, \sqrt {2} {\left ({\left (3 \, A + 7 \, B\right )} a^{2} c \cos \left (f x + e\right )^{2} - {\left (3 \, A + 7 \, B\right )} a^{2} c \cos \left (f x + e\right ) - 2 \, {\left (3 \, A + 7 \, B\right )} a^{2} c + {\left ({\left (3 \, A + 7 \, B\right )} a^{2} c \cos \left (f x + e\right ) + 2 \, {\left (3 \, A + 7 \, B\right )} a^{2} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (B a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, A + 10 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 6 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right ) + 3 \, {\left (A + B\right )} a^{2} + {\left (B a^{2} \cos \left (f x + e\right )^{2} - 3 \, {\left (A + 3 \, B\right )} a^{2} \cos \left (f x + e\right ) + 3 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(2)*((3*A + 7*B)*a^2*c*cos(f*x + e)^2 - (3*A + 7*B)*a^2*c*cos(f*x + e) - 2*(3*A + 7*B)*a^2*c + ((3*
A + 7*B)*a^2*c*cos(f*x + e) + 2*(3*A + 7*B)*a^2*c)*sin(f*x + e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin
(f*x + e) - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2
)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) - 4*(B*a^2*cos(f*x + e)^3 + (
3*A + 10*B)*a^2*cos(f*x + e)^2 + 6*(A + 2*B)*a^2*cos(f*x + e) + 3*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 3*(A +
3*B)*a^2*cos(f*x + e) + 3*(A + B)*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2*f
*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),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^2*sig
n(tan((f*x+exp(1))/2)-1)+240*B*a^2*sign(tan((f*x+exp(1))/2)-1))-1/144*(72*A*a^2*sign(tan((f*x+exp(1))/2)-1)+28
8*B*a^2*sign(tan((f*x+exp(1))/2)-1)))-1/144*(72*A*a^2*sign(tan((f*x+exp(1))/2)-1)+288*B*a^2*sign(tan((f*x+exp(
1))/2)-1)))-1/144*(72*A*a^2*sign(tan((f*x+exp(1))/2)-1)+240*B*a^2*sign(tan((f*x+exp(1))/2)-1)))+2*((-3*A*a^2*(
-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-3*B*a^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*t
an((f*x+exp(1))/2)^2+c))^3+A*a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-A*a^2*sqrt(c
)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+B*a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))-B*a^2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-A*a
^2*sqrt(c)*c-B*a^2*sqrt(c)*c)/(-(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2*sqrt(c)*(-s
qrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+c)^2/c/sign(tan((f*x+exp(1))/2)-1)+1/2*(-6*A*a^2-1
4*B*a^2)*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)
/sqrt(-c)/c/sign(tan((f*x+exp(1))/2)-1)))

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maple [A]  time = 1.46, size = 282, normalized size = 1.60 \[ \frac {a^{2} \left (\sin \left (f x +e \right ) \left (9 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-6 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+21 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-18 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-2 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}\right )-9 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+12 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-21 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+24 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+2 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{3 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*(sin(f*x+e)*(9*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2-6*A*(c+c*sin(f*x+e))^
(1/2)*c^(3/2)+21*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2-18*B*(c+c*sin(f*x+e))^(1/2)
*c^(3/2)-2*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2))-9*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c
^2+12*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-21*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+24
*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2)+2*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/c^(7/2)/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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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