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

Optimal. Leaf size=126 \[ -\frac {a (A-7 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}-\frac {a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/2*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^(5/2)-1/8*a*(A+9*B)*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(3/2)-1/16*a*(A-
7*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)

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Rubi [A]  time = 0.33, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2967, 2857, 2750, 2649, 206} \[ -\frac {a (A-7 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}-\frac {a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

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 2750

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

Rule 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx &=(a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}+\frac {a \int \frac {-A c-5 B c-4 B c \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c^2}\\ &=\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}-\frac {(a (A-7 B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 c^2}\\ &=\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac {(a (A-7 B)) \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^2 f}\\ &=-\frac {a (A-7 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^{5/2} f}+\frac {a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 2.24, size = 199, normalized size = 1.58 \[ -\frac {a (\sin (e+f x)-1) (\sin (e+f x)+1) \left (\frac {2 \sqrt {c} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) ((A+9 B) \sin (e+f x)+3 A-5 B)}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}+\sqrt {2} (A-7 B) \sec (e+f x) \sqrt {-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac {\sqrt {-c (\sin (e+f x)+1)}}{\sqrt {2} \sqrt {c}}\right )\right )}{16 c^{5/2} f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-1/16*(a*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])*(Sqrt[2]*(A - 7*B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[
2]*Sqrt[c])]*Sec[e + f*x]*Sqrt[-(c*(1 + Sin[e + f*x]))] + (2*Sqrt[c]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*
A - 5*B + (A + 9*B)*Sin[e + f*x]))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5))/(c^(5/2)*f*(Cos[(e + f*x)/2] + Si
n[(e + f*x)/2])^2*Sqrt[c - c*Sin[e + f*x]])

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fricas [B]  time = 0.46, size = 394, normalized size = 3.13 \[ -\frac {\sqrt {2} {\left ({\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{3} + 3 \, {\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{2} - 2 \, {\left (A - 7 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (A - 7 \, B\right )} a - {\left ({\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{2} - 2 \, {\left (A - 7 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (A - 7 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left ({\left (A + 9 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (3 \, A - 5 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (A + B\right )} a - {\left ({\left (A + 9 \, B\right )} a \cos \left (f x + e\right ) + 4 \, {\left (A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{32 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((A - 7*B)*a*cos(f*x + e)^3 + 3*(A - 7*B)*a*cos(f*x + e)^2 - 2*(A - 7*B)*a*cos(f*x + e) - 4*(A
- 7*B)*a - ((A - 7*B)*a*cos(f*x + e)^2 - 2*(A - 7*B)*a*cos(f*x + e) - 4*(A - 7*B)*a)*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*((A + 9*B)*a*cos(f*x + e)^2 - (3*A - 5*B)*a*cos(f*x + e) - 4*(A + B)*a - ((A + 9*B)*a*cos(f
*x + e) + 4*(A + B)*a)*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))*(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)2/f*2*(1/16*(-17*A*a*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan(
(f*x+exp(1))/2)^2+c))^7+7*B*a*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-23*A*a*sqrt(c)*
(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6+81*B*a*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+
sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-19*A*a*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+5
3*B*a*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+39*A*a*sqrt(c)*c*(-sqrt(c)*tan((f*x+e
xp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4-65*B*a*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+ex
p(1))/2)^2+c))^4+5*A*a*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+13*B*a*c^2*(-sqrt(
c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+7*A*a*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((
f*x+exp(1))/2)^2+c))-37*A*a*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-33*B*
a*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+19*B*a*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp
(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-3*A*a*sqrt(c)*c^3+5*B*a*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)+1/16*(-A*a+7*B*a)*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 = 1.61, size = 268, normalized size = 2.13 \[ \frac {a \left (-2 \sin \left (f x +e \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \left (A -7 B \right )-\sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \left (A -7 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-2 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}-4 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-14 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-18 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+28 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{16 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a*(-2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2*(A-7*B)-2^(1/2)*arctanh(
1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2*(A-7*B)*cos(f*x+e)^2+2*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^
(1/2)*2^(1/2)/c^(1/2))*c^2-2*A*(c+c*sin(f*x+e))^(3/2)*c^(1/2)-4*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-14*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))^(3/2)*c^(1/2)+28*B*(c+c*sin(f*x+
e))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2)/c^(9/2)/(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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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