3.314 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=283 \[ \frac{d^2 (15 A c-35 A d-51 B c+39 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{30 a^2 f}-\frac{(c-d)^2 (A (c+11 d)+3 B (c-5 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{d \left (15 A c^2-120 A c d+65 A d^2-99 B c^2+168 B c d-93 B d^2\right ) \cos (e+f x)}{15 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac{d (5 A-9 B) \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a \sin (e+f x)+a}} \]
[Out]
-((c - d)^2*(3*B*(c - 5*d) + A*(c + 11*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*(15*A*c^2 - 99*B*c^2 - 120*A*c*d + 168*B*c*d + 65*A*d^2 - 93*B*d^2)*Cos[e + f*x])/
(15*a*f*Sqrt[a + a*Sin[e + f*x]]) + (d^2*(15*A*c - 51*B*c - 35*A*d + 39*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f
*x]])/(30*a^2*f) + ((5*A - 9*B)*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(10*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A
- B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(2*f*(a + a*Sin[e + f*x])^(3/2))
________________________________________________________________________________________
Rubi [A] time = 0.999728, antiderivative size = 283, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used =
{2977, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac{d^2 (15 A c-35 A d-51 B c+39 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{30 a^2 f}-\frac{(c-d)^2 (A (c+11 d)+3 B (c-5 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{d \left (15 A c^2-120 A c d+65 A d^2-99 B c^2+168 B c d-93 B d^2\right ) \cos (e+f x)}{15 a f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac{d (5 A-9 B) \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a \sin (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
-((c - d)^2*(3*B*(c - 5*d) + A*(c + 11*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*(15*A*c^2 - 99*B*c^2 - 120*A*c*d + 168*B*c*d + 65*A*d^2 - 93*B*d^2)*Cos[e + f*x])/
(15*a*f*Sqrt[a + a*Sin[e + f*x]]) + (d^2*(15*A*c - 51*B*c - 35*A*d + 39*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f
*x]])/(30*a^2*f) + ((5*A - 9*B)*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(10*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A
- B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(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 2983
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 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))^3}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{(c+d \sin (e+f x))^2 \left (\frac{1}{2} a (3 B (c-2 d)+A (c+6 d))-\frac{1}{2} a (5 A-9 B) d \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{(c+d \sin (e+f x)) \left (\frac{1}{4} a^2 \left (5 A \left (c^2+7 c d-4 d^2\right )+3 B \left (5 c^2-13 c d+12 d^2\right )\right )-\frac{1}{4} a^2 d (15 A c-51 B c-35 A d+39 B d) \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{5 a^3}\\ &=\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 c \left (5 A \left (c^2+7 c d-4 d^2\right )+3 B \left (5 c^2-13 c d+12 d^2\right )\right )+\left (-\frac{1}{4} a^2 c d (15 A c-51 B c-35 A d+39 B d)+\frac{1}{4} a^2 d \left (5 A \left (c^2+7 c d-4 d^2\right )+3 B \left (5 c^2-13 c d+12 d^2\right )\right )\right ) \sin (e+f x)-\frac{1}{4} a^2 d^2 (15 A c-51 B c-35 A d+39 B d) \sin ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{5 a^3}\\ &=\frac{d^2 (15 A c-51 B c-35 A d+39 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{30 a^2 f}+\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{2 \int \frac{\frac{1}{8} a^3 \left (3 B \left (15 c^3-39 c^2 d+53 c d^2-13 d^3\right )+5 A \left (3 c^3+21 c^2 d-15 c d^2+7 d^3\right )\right )-\frac{1}{4} a^3 d \left (15 A c^2-99 B c^2-120 A c d+168 B c d+65 A d^2-93 B d^2\right ) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{15 a^4}\\ &=\frac{d \left (15 A c^2-99 B c^2-120 A c d+168 B c d+65 A d^2-93 B d^2\right ) \cos (e+f x)}{15 a f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (15 A c-51 B c-35 A d+39 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{30 a^2 f}+\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\left ((c-d)^2 (3 B (c-5 d)+A (c+11 d))\right ) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a}\\ &=\frac{d \left (15 A c^2-99 B c^2-120 A c d+168 B c d+65 A d^2-93 B d^2\right ) \cos (e+f x)}{15 a f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (15 A c-51 B c-35 A d+39 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{30 a^2 f}+\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left ((c-d)^2 (3 B (c-5 d)+A (c+11 d))\right ) \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)^2 (3 B (c-5 d)+A (c+11 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 \left (15 A c^2-99 B c^2-120 A c d+168 B c d+65 A d^2-93 B d^2\right ) \cos (e+f x)}{15 a f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (15 A c-51 B c-35 A d+39 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{30 a^2 f}+\frac{(5 A-9 B) d \cos (e+f x) (c+d \sin (e+f x))^2}{10 a f \sqrt{a+a \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 1.06042, size = 684, normalized size = 2.42 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((30+30 i) (-1)^{3/4} (c-d)^2 (A (c+11 d)+3 B (c-5 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 )-90 A c^2 d \sin \left (\frac{1}{2} (e+f x)\right )+90 A c^2 d \cos \left (\frac{1}{2} (e+f x)\right )+30 A c^3 \sin \left (\frac{1}{2} (e+f x)\right )-30 A c^3 \cos \left (\frac{1}{2} (e+f x)\right )+270 A c d^2 \sin \left (\frac{1}{2} (e+f x)\right )-180 A c d^2 \sin \left (\frac{3}{2} (e+f x)\right )-270 A c d^2 \cos \left (\frac{1}{2} (e+f x)\right )-180 A c d^2 \cos \left (\frac{3}{2} (e+f x)\right )-110 A d^3 \sin \left (\frac{1}{2} (e+f x)\right )+70 A d^3 \sin \left (\frac{3}{2} (e+f x)\right )-10 A d^3 \sin \left (\frac{5}{2} (e+f x)\right )+110 A d^3 \cos \left (\frac{1}{2} (e+f x)\right )+70 A d^3 \cos \left (\frac{3}{2} (e+f x)\right )+10 A d^3 \cos \left (\frac{5}{2} (e+f x)\right )+270 B c^2 d \sin \left (\frac{1}{2} (e+f x)\right )-180 B c^2 d \sin \left (\frac{3}{2} (e+f x)\right )-270 B c^2 d \cos \left (\frac{1}{2} (e+f x)\right )-180 B c^2 d \cos \left (\frac{3}{2} (e+f x)\right )-30 B c^3 \sin \left (\frac{1}{2} (e+f x)\right )+30 B c^3 \cos \left (\frac{1}{2} (e+f x)\right )-330 B c d^2 \sin \left (\frac{1}{2} (e+f x)\right )+210 B c d^2 \sin \left (\frac{3}{2} (e+f x)\right )-30 B c d^2 \sin \left (\frac{5}{2} (e+f x)\right )+330 B c d^2 \cos \left (\frac{1}{2} (e+f x)\right )+210 B c d^2 \cos \left (\frac{3}{2} (e+f x)\right )+30 B c d^2 \cos \left (\frac{5}{2} (e+f x)\right )+165 B d^3 \sin \left (\frac{1}{2} (e+f x)\right )-123 B d^3 \sin \left (\frac{3}{2} (e+f x)\right )+9 B d^3 \sin \left (\frac{5}{2} (e+f x)\right )+3 B d^3 \sin \left (\frac{7}{2} (e+f x)\right )-165 B d^3 \cos \left (\frac{1}{2} (e+f x)\right )-123 B d^3 \cos \left (\frac{3}{2} (e+f x)\right )-9 B d^3 \cos \left (\frac{5}{2} (e+f x)\right )+3 B d^3 \cos \left (\frac{7}{2} (e+f x)\right )\right )}{60 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])^3)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-30*A*c^3*Cos[(e + f*x)/2] + 30*B*c^3*Cos[(e + f*x)/2] + 90*A*c^2*d*Co
s[(e + f*x)/2] - 270*B*c^2*d*Cos[(e + f*x)/2] - 270*A*c*d^2*Cos[(e + f*x)/2] + 330*B*c*d^2*Cos[(e + f*x)/2] +
110*A*d^3*Cos[(e + f*x)/2] - 165*B*d^3*Cos[(e + f*x)/2] - 180*B*c^2*d*Cos[(3*(e + f*x))/2] - 180*A*c*d^2*Cos[(
3*(e + f*x))/2] + 210*B*c*d^2*Cos[(3*(e + f*x))/2] + 70*A*d^3*Cos[(3*(e + f*x))/2] - 123*B*d^3*Cos[(3*(e + f*x
))/2] + 30*B*c*d^2*Cos[(5*(e + f*x))/2] + 10*A*d^3*Cos[(5*(e + f*x))/2] - 9*B*d^3*Cos[(5*(e + f*x))/2] + 3*B*d
^3*Cos[(7*(e + f*x))/2] + 30*A*c^3*Sin[(e + f*x)/2] - 30*B*c^3*Sin[(e + f*x)/2] - 90*A*c^2*d*Sin[(e + f*x)/2]
+ 270*B*c^2*d*Sin[(e + f*x)/2] + 270*A*c*d^2*Sin[(e + f*x)/2] - 330*B*c*d^2*Sin[(e + f*x)/2] - 110*A*d^3*Sin[(
e + f*x)/2] + 165*B*d^3*Sin[(e + f*x)/2] + (30 + 30*I)*(-1)^(3/4)*(c - d)^2*(3*B*(c - 5*d) + A*(c + 11*d))*Arc
Tanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 180*B*c^2*d*Sin
[(3*(e + f*x))/2] - 180*A*c*d^2*Sin[(3*(e + f*x))/2] + 210*B*c*d^2*Sin[(3*(e + f*x))/2] + 70*A*d^3*Sin[(3*(e +
f*x))/2] - 123*B*d^3*Sin[(3*(e + f*x))/2] - 30*B*c*d^2*Sin[(5*(e + f*x))/2] - 10*A*d^3*Sin[(5*(e + f*x))/2] +
9*B*d^3*Sin[(5*(e + f*x))/2] + 3*B*d^3*Sin[(7*(e + f*x))/2]))/(60*f*(a*(1 + Sin[e + f*x]))^(3/2))
________________________________________________________________________________________
Maple [B] time = 1.46, size = 1030, normalized size = 3.6 \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))^3/(a+a*sin(f*x+e))^(3/2),x)
[Out]
-1/60*(sin(f*x+e)*(-40*A*(a-a*sin(f*x+e))^(3/2)*a^(3/2)*d^3+360*A*c*d^2*a^(5/2)*(a-a*sin(f*x+e))^(1/2)-120*A*a
^(5/2)*d^3*(a-a*sin(f*x+e))^(1/2)+24*B*d^3*(a-a*sin(f*x+e))^(5/2)*a^(1/2)-120*B*(a-a*sin(f*x+e))^(3/2)*a^(3/2)
*c*d^2+360*B*c^2*d*a^(5/2)*(a-a*sin(f*x+e))^(1/2)-360*B*a^(5/2)*c*d^2*(a-a*sin(f*x+e))^(1/2)+240*B*a^(5/2)*d^3
*(a-a*sin(f*x+e))^(1/2)+15*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c^3+135*A*2^(1/2)
*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c^2*d-315*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*a^3*c*d^2+165*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*d^3+45*B*2^
(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c^3-315*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(
1/2)*2^(1/2)/a^(1/2))*a^3*c^2*d+495*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c*d^2-22
5*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*d^3)-40*A*(a-a*sin(f*x+e))^(3/2)*a^(3/2)*d
^3+30*A*(a-a*sin(f*x+e))^(1/2)*a^(5/2)*c^3-90*A*(a-a*sin(f*x+e))^(1/2)*a^(5/2)*c^2*d+450*A*c*d^2*a^(5/2)*(a-a*
sin(f*x+e))^(1/2)-150*A*a^(5/2)*d^3*(a-a*sin(f*x+e))^(1/2)+24*B*d^3*(a-a*sin(f*x+e))^(5/2)*a^(1/2)-120*B*(a-a*
sin(f*x+e))^(3/2)*a^(3/2)*c*d^2-30*B*(a-a*sin(f*x+e))^(1/2)*a^(5/2)*c^3+450*B*c^2*d*a^(5/2)*(a-a*sin(f*x+e))^(
1/2)-450*B*a^(5/2)*c*d^2*(a-a*sin(f*x+e))^(1/2)+270*B*a^(5/2)*d^3*(a-a*sin(f*x+e))^(1/2)+15*A*2^(1/2)*arctanh(
1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c^3+135*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a
^(1/2))*a^3*c^2*d-315*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c*d^2+165*A*2^(1/2)*ar
ctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*d^3+45*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1
/2)/a^(1/2))*a^3*c^3-315*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c^2*d+495*B*2^(1/2)
*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c*d^2-225*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*a^3*d^3)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2)/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 )}^{3}}{{\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))^3/(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)^3/(a*sin(f*x + e) + a)^(3/2), x)
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Fricas [B] time = 2.03246, size = 1901, normalized size = 6.72 \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))^3/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
1/120*(15*sqrt(2)*(2*(A + 3*B)*c^3 + 6*(3*A - 7*B)*c^2*d - 6*(7*A - 11*B)*c*d^2 + 2*(11*A - 15*B)*d^3 - ((A +
3*B)*c^3 + 3*(3*A - 7*B)*c^2*d - 3*(7*A - 11*B)*c*d^2 + (11*A - 15*B)*d^3)*cos(f*x + e)^2 + ((A + 3*B)*c^3 + 3
*(3*A - 7*B)*c^2*d - 3*(7*A - 11*B)*c*d^2 + (11*A - 15*B)*d^3)*cos(f*x + e) + (2*(A + 3*B)*c^3 + 6*(3*A - 7*B)
*c^2*d - 6*(7*A - 11*B)*c*d^2 + 2*(11*A - 15*B)*d^3 + ((A + 3*B)*c^3 + 3*(3*A - 7*B)*c^2*d - 3*(7*A - 11*B)*c*
d^2 + (11*A - 15*B)*d^3)*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)*(cos(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*(12*B*d^3*cos(f*x + e)^4 - 1
5*(A - B)*c^3 + 45*(A - B)*c^2*d - 45*(A - B)*c*d^2 + 15*(A - B)*d^3 + 4*(15*B*c*d^2 + (5*A - 3*B)*d^3)*cos(f*
x + e)^3 - 4*(45*B*c^2*d + 15*(3*A - 4*B)*c*d^2 - 4*(5*A - 9*B)*d^3)*cos(f*x + e)^2 - 15*((A - B)*c^3 - 3*(A -
5*B)*c^2*d + 15*(A - B)*c*d^2 - (5*A - 9*B)*d^3)*cos(f*x + e) + (12*B*d^3*cos(f*x + e)^3 + 15*(A - B)*c^3 - 4
5*(A - B)*c^2*d + 45*(A - B)*c*d^2 - 15*(A - B)*d^3 - 4*(15*B*c*d^2 + (5*A - 6*B)*d^3)*cos(f*x + e)^2 - 60*(3*
B*c^2*d + 3*(A - B)*c*d^2 - (A - 2*B)*d^3)*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))**3/(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))^3/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
sage2