3.315 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=203 \[ -\frac{(c-d) (A c+7 A d+3 B c-11 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{d^2 (3 A-7 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{6 a^2 f}+\frac{d (3 A c-9 A d-15 B c+13 B d) \cos (e+f x)}{3 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-((c - d)*(A*c + 3*B*c + 7*A*d - 11*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(
2*Sqrt[2]*a^(3/2)*f) + (d*(3*A*c - 15*B*c - 9*A*d + 13*B*d)*Cos[e + f*x])/(3*a*f*Sqrt[a + a*Sin[e + f*x]]) + (
(3*A - 7*B)*d^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(6*a^2*f) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^
2)/(2*f*(a + a*Sin[e + f*x])^(3/2))
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Rubi [A] time = 0.575366, antiderivative size = 203, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{2977, 2968, 3023, 2751, 2649, 206} \[ -\frac{(c-d) (A c+7 A d+3 B c-11 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{d^2 (3 A-7 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{6 a^2 f}+\frac{d (3 A c-9 A d-15 B c+13 B d) \cos (e+f x)}{3 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
-((c - d)*(A*c + 3*B*c + 7*A*d - 11*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(
2*Sqrt[2]*a^(3/2)*f) + (d*(3*A*c - 15*B*c - 9*A*d + 13*B*d)*Cos[e + f*x])/(3*a*f*Sqrt[a + a*Sin[e + f*x]]) + (
(3*A - 7*B)*d^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(6*a^2*f) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^
2)/(2*f*(a + a*Sin[e + f*x])^(3/2))
Rule 2977
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 2751
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m,
-2^(-1)]
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+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{(c+d \sin (e+f x)) \left (\frac{1}{2} a (A c+3 B c+4 A d-4 B d)-\frac{1}{2} a (3 A-7 B) d \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\frac{1}{2} a c (A c+3 B c+4 A d-4 B d)+\left (-\frac{1}{2} a (3 A-7 B) c d+\frac{1}{2} a d (A c+3 B c+4 A d-4 B d)\right ) \sin (e+f x)-\frac{1}{2} a (3 A-7 B) d^2 \sin ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=\frac{(3 A-7 B) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{6 a^2 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{-\frac{1}{4} a^2 \left ((3 A-7 B) d^2-3 c (A c+3 B c+4 A d-4 B d)\right )-\frac{1}{2} a^2 d (3 A c-15 B c-9 A d+13 B d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{3 a^3}\\ &=\frac{d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt{a+a \sin (e+f x)}}+\frac{(3 A-7 B) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{6 a^2 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{((c-d) (A c+3 B c+7 A d-11 B d)) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a}\\ &=\frac{d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt{a+a \sin (e+f x)}}+\frac{(3 A-7 B) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{6 a^2 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{((c-d) (A c+3 B c+7 A d-11 B d)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{2 a f}\\ &=-\frac{(c-d) (A c+3 B c+7 A d-11 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt{a+a \sin (e+f x)}}+\frac{(3 A-7 B) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{6 a^2 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.734585, size = 357, normalized size = 1.76 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (6 (A-B) (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-3 (A-B) (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+6 d (-2 A d-4 B c+3 B d) \cos \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-6 d (-2 A d-4 B c+3 B d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(3+3 i) (-1)^{3/4} (c-d) (A c+7 A d+3 B c-11 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-2 B d^2 \cos \left (\frac{3}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-2 B d^2 \sin \left (\frac{3}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2\right )}{6 f (a (\sin (e+f x)+1))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(A - B)*(c - d)^2*Sin[(e + f*x)/2] - 3*(A - B)*(c - d)^2*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2]) + (3 + 3*I)*(-1)^(3/4)*(c - d)*(A*c + 3*B*c + 7*A*d - 11*B*d)*ArcTanh[(1/2 + I/2)*(
-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 6*d*(-4*B*c - 2*A*d + 3*B*d)*Cos[
(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*B*d^2*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^2 - 6*d*(-4*B*c - 2*A*d + 3*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*B*d^
2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sin[(3*(e + f*x))/2]))/(6*f*(a*(1 + Sin[e + f*x]))^(3/2))
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Maple [B] time = 1.345, size = 694, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x)
[Out]
-1/12/a^(7/2)*(sin(f*x+e)*(3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+18*A*2^(1/2
)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-21*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*
2^(1/2)/a^(1/2))*a^2*d^2+24*A*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+9*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*a^2*c^2-42*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+33*B*2^(1/
2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-8*B*d^2*(a-a*sin(f*x+e))^(3/2)*a^(1/2)+48*B*c*d
*a^(3/2)*(a-a*sin(f*x+e))^(1/2)-24*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2))+3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+18*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-21
*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+6*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2-
12*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d+30*A*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+9*B*2^(1/2)*arctanh(1/2*(a-a*s
in(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-42*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2
*c*d+33*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-8*B*d^2*(a-a*sin(f*x+e))^(3/2)*a
^(1/2)-6*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2+60*B*c*d*a^(3/2)*(a-a*sin(f*x+e))^(1/2)-30*B*d^2*a^(3/2)*(a-a*si
n(f*x+e))^(1/2))*(-a*(-1+sin(f*x+e)))^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a)^(3/2), x)
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Fricas [B] time = 1.8714, size = 1422, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
-1/24*(3*sqrt(2)*(2*(A + 3*B)*c^2 + 4*(3*A - 7*B)*c*d - 2*(7*A - 11*B)*d^2 - ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*
d - (7*A - 11*B)*d^2)*cos(f*x + e)^2 + ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*d - (7*A - 11*B)*d^2)*cos(f*x + e) + (
2*(A + 3*B)*c^2 + 4*(3*A - 7*B)*c*d - 2*(7*A - 11*B)*d^2 + ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*d - (7*A - 11*B)*d
^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(c
os(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^
2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(4*B*d^2*cos(f*x + e)^3 - 3*(A - B)*c^2 + 6*(A -
B)*c*d - 3*(A - B)*d^2 - 4*(6*B*c*d + (3*A - 4*B)*d^2)*cos(f*x + e)^2 - 3*((A - B)*c^2 - 2*(A - 5*B)*c*d + 5*(
A - B)*d^2)*cos(f*x + e) - (4*B*d^2*cos(f*x + e)^2 - 3*(A - B)*c^2 + 6*(A - B)*c*d - 3*(A - B)*d^2 + 12*(2*B*c
*d + (A - B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x
+ e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
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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+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(3/2),x)
[Out]
Timed out
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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+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
sage2