3.321 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=308 \[ \frac{d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt{a \sin (e+f x)+a}}-\frac{(c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{d^2 (9 A c+39 A d+15 B c-95 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{48 a^3 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{(3 A c+9 A d+5 B c-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-((c - d)*(B*(5*c^2 + 62*c*d - 163*d^2) + 3*A*(c^2 + 6*c*d + 25*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*
Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) + (d*(A*(9*c^2 + 36*c*d - 93*d^2) + B*(15*c^2 - 228*c*d + 1
97*d^2))*Cos[e + f*x])/(24*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + (d^2*(9*A*c + 15*B*c + 39*A*d - 95*B*d)*Cos[e + f
*x]*Sqrt[a + a*Sin[e + f*x]])/(48*a^3*f) - ((3*A*c + 5*B*c + 9*A*d - 17*B*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])
^2)/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*f*(a + a*Sin[e + f*
x])^(5/2))
________________________________________________________________________________________
Rubi [A] time = 1.05919, antiderivative size = 308, normalized size of antiderivative = 1.,
number of steps used = 7, 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} \[ \frac{d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt{a \sin (e+f x)+a}}-\frac{(c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{d^2 (9 A c+39 A d+15 B c-95 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{48 a^3 f}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{(3 A c+9 A d+5 B c-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3)/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
-((c - d)*(B*(5*c^2 + 62*c*d - 163*d^2) + 3*A*(c^2 + 6*c*d + 25*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*
Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) + (d*(A*(9*c^2 + 36*c*d - 93*d^2) + B*(15*c^2 - 228*c*d + 1
97*d^2))*Cos[e + f*x])/(24*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + (d^2*(9*A*c + 15*B*c + 39*A*d - 95*B*d)*Cos[e + f
*x]*Sqrt[a + a*Sin[e + f*x]])/(48*a^3*f) - ((3*A*c + 5*B*c + 9*A*d - 17*B*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])
^2)/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*f*(a + a*Sin[e + f*
x])^(5/2))
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))^3}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\int \frac{(c+d \sin (e+f x))^2 \left (\frac{1}{2} a (3 A c+5 B c+6 A d-6 B d)-\frac{1}{2} a (3 A-11 B) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\int \frac{(c+d \sin (e+f x)) \left (\frac{1}{4} a^2 \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )-\frac{1}{4} a^2 d (9 A c+15 B c+39 A d-95 B d) \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{8 a^4}\\ &=-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\int \frac{\frac{1}{4} a^2 c \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )+\left (-\frac{1}{4} a^2 c d (9 A c+15 B c+39 A d-95 B d)+\frac{1}{4} a^2 d \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )\right ) \sin (e+f x)-\frac{1}{4} a^2 d^2 (9 A c+15 B c+39 A d-95 B d) \sin ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{8 a^4}\\ &=\frac{d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{48 a^3 f}-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\int \frac{\frac{1}{8} a^3 \left (3 A \left (3 c^3+9 c^2 d+33 c d^2-13 d^3\right )+B \left (15 c^3+141 c^2 d-219 c d^2+95 d^3\right )\right )-\frac{1}{4} a^3 d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{12 a^5}\\ &=\frac{d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{48 a^3 f}-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\left ((c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right )\right ) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=\frac{d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{48 a^3 f}-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{\left ((c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right )\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 )}{16 a^2 f}\\ &=-\frac{(c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{48 a^3 f}-\frac{(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end{align*}
Mathematica [C] time = 1.78719, size = 523, normalized size = 1.7 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((3+3 i) (-1)^{3/4} (c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+(24+24 i) d^2 (-2 A d-6 B c+5 B d) \left (\cos \left (\frac{1}{2} (e+f x)\right )+i \sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+(24+24 i) d^2 (2 A d+6 B c-5 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+i \cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+24 (A-B) (c-d)^3 \sin \left (\frac{1}{2} (e+f x)\right )-3 (c-d)^2 (3 A (c+7 d)+B (5 c-29 d)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+6 (c-d)^2 (3 A (c+7 d)+B (5 c-29 d)) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-12 (A-B) (c-d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-16 B d^3 \cos \left (\frac{3}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4-16 B d^3 \sin \left (\frac{3}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{48 f (a (\sin (e+f x)+1))^{5/2}} \]
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])^(5/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(24*(A - B)*(c - d)^3*Sin[(e + f*x)/2] - 12*(A - B)*(c - d)^3*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]) + 6*(c - d)^2*(B*(5*c - 29*d) + 3*A*(c + 7*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^2 - 3*(c - d)^2*(B*(5*c - 29*d) + 3*A*(c + 7*d))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3
+ (3 + 3*I)*(-1)^(3/4)*(c - d)*(B*(5*c^2 + 62*c*d - 163*d^2) + 3*A*(c^2 + 6*c*d + 25*d^2))*ArcTanh[(1/2 + I/2)
*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 16*B*d^3*Cos[(3*(e + f*x))/2]*(
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (24 + 24*I)*d^2*(-6*B*c - 2*A*d + 5*B*d)*(Cos[(e + f*x)/2] + I*Sin[(e
+ f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (24 + 24*I)*d^2*(6*B*c + 2*A*d - 5*B*d)*(I*Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 16*B*d^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2
])^4*Sin[(3*(e + f*x))/2]))/(48*f*(a*(1 + Sin[e + f*x]))^(5/2))
________________________________________________________________________________________
Maple [B] time = 2.148, size = 1438, normalized size = 4.7 \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))^(5/2),x)
[Out]
-1/96*(60*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^3+36*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^3-18*A*(a-a*sin(f*x+e))^(
3/2)*a^(1/2)*c^3-126*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^3-30*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^3+342*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^2+342*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*d-1350*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^2+90*
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*d+2*sin(f*x+e)*(9*A*2^(1/2)*arctanh(1/2*
(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+45*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*d+171*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^2-225*A*2^(1/2)*arctanh
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3+192*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3+15*B*2^(1/2)*arc
tanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+171*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*d-675*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^2+489*B*2^(1/
2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3+576*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d^2-384*
B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3-64*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^3)+(-9*A*2^(1/2)*arctanh(1/2*(a-a*s
in(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3-45*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*d-171*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^2+225*A*2^(1/2)*arctanh(1/2*(
a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3-192*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3-15*B*2^(1/2)*arctanh(1
/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3-171*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*d+675*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^2-489*B*2^(1/2)*arc
tanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3+64*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^3-576*B*(a-a*si
n(f*x+e))^(1/2)*a^(3/2)*c*d^2+384*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3)*cos(f*x+e)^2+234*B*(a-a*sin(f*x+e))^(3
/2)*a^(1/2)*c^2*d-378*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d^2+108*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2*d-396*A*
(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d^2-396*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2*d+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^3-450*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a
^2*d^3+30*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+978*B*2^(1/2)*arctanh(1/2*(a-a
*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3+1836*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d^2-90*A*(a-a*sin(f*x+e))^
(3/2)*a^(1/2)*c^2*d+234*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d^2+612*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3-1092*B
*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^3+46*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^3)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2
)/(1+sin(f*x+e))/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{5}{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))^(5/2),x, algorithm="maxima")
[Out]
integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^3/(a*sin(f*x + e) + a)^(5/2), x)
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Fricas [B] time = 2.39947, size = 2394, normalized size = 7.77 \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))^(5/2),x, algorithm="fricas")
[Out]
-1/192*(3*sqrt(2)*(4*(3*A + 5*B)*c^3 + 12*(5*A + 19*B)*c^2*d + 12*(19*A - 75*B)*c*d^2 - 4*(75*A - 163*B)*d^3 -
((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^3 - 3*((3*
A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^2 + 2*((3*A + 5
*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e) + (4*(3*A + 5*B)*c^3
+ 12*(5*A + 19*B)*c^2*d + 12*(19*A - 75*B)*c*d^2 - 4*(75*A - 163*B)*d^3 - ((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c
^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^2 + 2*((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d
+ 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*s
qrt(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*(32*B*d
^3*cos(f*x + e)^4 - 12*(A - B)*c^3 + 36*(A - B)*c^2*d - 36*(A - B)*c*d^2 + 12*(A - B)*d^3 + 32*(9*B*c*d^2 + (3
*A - 5*B)*d^3)*cos(f*x + e)^3 - 3*((3*A + 5*B)*c^3 + 3*(5*A - 13*B)*c^2*d - 3*(13*A - 53*B)*c*d^2 + (53*A - 93
*B)*d^3)*cos(f*x + e)^2 - 3*((7*A + B)*c^3 + 3*(A - 9*B)*c^2*d - 27*(A - 9*B)*c*d^2 + 9*(9*A - 17*B)*d^3)*cos(
f*x + e) + (32*B*d^3*cos(f*x + e)^3 + 12*(A - B)*c^3 - 36*(A - B)*c^2*d + 36*(A - B)*c*d^2 - 12*(A - B)*d^3 -
96*(3*B*c*d^2 + (A - 2*B)*d^3)*cos(f*x + e)^2 - 3*((3*A + 5*B)*c^3 + 3*(5*A - 13*B)*c^2*d - 3*(13*A - 85*B)*c*
d^2 + (85*A - 157*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*
f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*si
n(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))**(5/2),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))^(5/2),x, algorithm="giac")
[Out]
sage2