3.322 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=219 \[ -\frac {\left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (A-9 B) \cos (e+f x)}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {(c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-1/16*(c-d)*(3*A*c+5*A*d+5*B*c-13*B*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(A-B)*cos(f*x+e)*(c+d*sin(f*x
+e))^2/f/(a+a*sin(f*x+e))^(5/2)-1/32*(B*(5*c^2+38*c*d-75*d^2)+A*(3*c^2+10*c*d+19*d^2))*arctanh(1/2*cos(f*x+e)*
a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)+1/4*(A-9*B)*d^2*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1
/2)
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Rubi [A] time = 0.58, antiderivative size = 219, normalized size of antiderivative = 1.00,
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, 3019, 2751, 2649, 206} \[ -\frac {\left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (A-9 B) \cos (e+f x)}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {(c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
-((B*(5*c^2 + 38*c*d - 75*d^2) + A*(3*c^2 + 10*c*d + 19*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) - ((c - d)*(3*A*c + 5*B*c + 5*A*d - 13*B*d)*Cos[e + f*x])/(16*a*f*(
a + a*Sin[e + f*x])^(3/2)) + ((A - 9*B)*d^2*Cos[e + f*x])/(4*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e
+ f*x]*(c + d*Sin[e + f*x])^2)/(4*f*(a + a*Sin[e + f*x])^(5/2))
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 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 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 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 3019
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 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))^{5/2}} \, dx &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {(c+d \sin (e+f x)) \left (\frac {1}{2} a (3 A c+5 B c+4 A d-4 B d)-\frac {1}{2} a (A-9 B) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a c (3 A c+5 B c+4 A d-4 B d)+\left (-\frac {1}{2} a (A-9 B) c d+\frac {1}{2} a d (3 A c+5 B c+4 A d-4 B d)\right ) \sin (e+f x)-\frac {1}{2} a (A-9 B) d^2 \sin ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{4} a^2 \left (B \left (5 c^2+38 c d-39 d^2\right )+A \left (3 c^2+10 c d+15 d^2\right )\right )+a^2 (A-9 B) d^2 \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{8 a^4}\\ &=-\frac {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(A-9 B) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\right )\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(A-9 B) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\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 {\left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 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 {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(A-9 B) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end {align*}
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Mathematica [C] time = 1.14, size = 544, normalized size = 2.48 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left ((2+2 i) (-1)^{3/4} \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+11 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 A c^2 \sin \left (\frac {3}{2} (e+f x)\right )-11 A c^2 \cos \left (\frac {1}{2} (e+f x)\right )-3 A c^2 \cos \left (\frac {3}{2} (e+f x)\right )-6 A c d \sin \left (\frac {1}{2} (e+f x)\right )-10 A c d \sin \left (\frac {3}{2} (e+f x)\right )+6 A c d \cos \left (\frac {1}{2} (e+f x)\right )-10 A c d \cos \left (\frac {3}{2} (e+f x)\right )-5 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )+13 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )+5 A d^2 \cos \left (\frac {1}{2} (e+f x)\right )+13 A d^2 \cos \left (\frac {3}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-5 B c^2 \sin \left (\frac {3}{2} (e+f x)\right )+3 B c^2 \cos \left (\frac {1}{2} (e+f x)\right )-5 B c^2 \cos \left (\frac {3}{2} (e+f x)\right )-10 B c d \sin \left (\frac {1}{2} (e+f x)\right )+26 B c d \sin \left (\frac {3}{2} (e+f x)\right )+10 B c d \cos \left (\frac {1}{2} (e+f x)\right )+26 B c d \cos \left (\frac {3}{2} (e+f x)\right )+45 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )-69 B d^2 \sin \left (\frac {3}{2} (e+f x)\right )-16 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )-45 B d^2 \cos \left (\frac {1}{2} (e+f x)\right )-69 B d^2 \cos \left (\frac {3}{2} (e+f x)\right )+16 B d^2 \cos \left (\frac {5}{2} (e+f x)\right )\right )}{32 f (a (\sin (e+f x)+1))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-11*A*c^2*Cos[(e + f*x)/2] + 3*B*c^2*Cos[(e + f*x)/2] + 6*A*c*d*Cos[(e
+ f*x)/2] + 10*B*c*d*Cos[(e + f*x)/2] + 5*A*d^2*Cos[(e + f*x)/2] - 45*B*d^2*Cos[(e + f*x)/2] - 3*A*c^2*Cos[(3
*(e + f*x))/2] - 5*B*c^2*Cos[(3*(e + f*x))/2] - 10*A*c*d*Cos[(3*(e + f*x))/2] + 26*B*c*d*Cos[(3*(e + f*x))/2]
+ 13*A*d^2*Cos[(3*(e + f*x))/2] - 69*B*d^2*Cos[(3*(e + f*x))/2] + 16*B*d^2*Cos[(5*(e + f*x))/2] + 11*A*c^2*Sin
[(e + f*x)/2] - 3*B*c^2*Sin[(e + f*x)/2] - 6*A*c*d*Sin[(e + f*x)/2] - 10*B*c*d*Sin[(e + f*x)/2] - 5*A*d^2*Sin[
(e + f*x)/2] + 45*B*d^2*Sin[(e + f*x)/2] + (2 + 2*I)*(-1)^(3/4)*(B*(5*c^2 + 38*c*d - 75*d^2) + A*(3*c^2 + 10*c
*d + 19*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
- 3*A*c^2*Sin[(3*(e + f*x))/2] - 5*B*c^2*Sin[(3*(e + f*x))/2] - 10*A*c*d*Sin[(3*(e + f*x))/2] + 26*B*c*d*Sin[(
3*(e + f*x))/2] + 13*A*d^2*Sin[(3*(e + f*x))/2] - 69*B*d^2*Sin[(3*(e + f*x))/2] - 16*B*d^2*Sin[(5*(e + f*x))/2
]))/(32*f*(a*(1 + Sin[e + f*x]))^(5/2))
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fricas [B] time = 0.48, size = 744, normalized size = 3.40 \[ -\frac {\sqrt {2} {\left ({\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 4 \, {\left (3 \, A + 5 \, B\right )} c^{2} - 8 \, {\left (5 \, A + 19 \, B\right )} c d - 4 \, {\left (19 \, A - 75 \, B\right )} d^{2} + 3 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (3 \, A + 5 \, B\right )} c^{2} + 8 \, {\left (5 \, A + 19 \, B\right )} c d + 4 \, {\left (19 \, A - 75 \, B\right )} d^{2} - {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 (32 \, B d^{2} \cos \left (f x + e\right )^{3} - 4 \, {\left (A - B\right )} c^{2} + 8 \, {\left (A - B\right )} c d - 4 \, {\left (A - B\right )} d^{2} - {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A - 13 \, B\right )} c d - {\left (13 \, A - 53 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left ({\left (7 \, A + B\right )} c^{2} + 2 \, {\left (A - 9 \, B\right )} c d - 9 \, {\left (A - 9 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (32 \, B d^{2} \cos \left (f x + e\right )^{2} - 4 \, {\left (A - B\right )} c^{2} + 8 \, {\left (A - B\right )} c d - 4 \, {\left (A - B\right )} d^{2} + {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A - 13 \, B\right )} c d - {\left (13 \, A - 85 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
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))^(5/2),x, algorithm="fricas")
[Out]
-1/64*(sqrt(2)*(((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f*x + e)^3 - 4*(3*A + 5*B)*c^2
- 8*(5*A + 19*B)*c*d - 4*(19*A - 75*B)*d^2 + 3*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(
f*x + e)^2 - 2*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f*x + e) - (4*(3*A + 5*B)*c^2 +
8*(5*A + 19*B)*c*d + 4*(19*A - 75*B)*d^2 - ((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f*x
+ e)^2 + 2*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*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)*(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^2*cos(f*x + e)^3 - 4*(A - B)*c^2 + 8*(A - B)*c*d - 4*(A - B)*d^2 - ((3*A + 5*B)*c^2 +
2*(5*A - 13*B)*c*d - (13*A - 53*B)*d^2)*cos(f*x + e)^2 - ((7*A + B)*c^2 + 2*(A - 9*B)*c*d - 9*(A - 9*B)*d^2)*
cos(f*x + e) - (32*B*d^2*cos(f*x + e)^2 - 4*(A - B)*c^2 + 8*(A - B)*c*d - 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 + 2
*(5*A - 13*B)*c*d - (13*A - 85*B)*d^2)*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)*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+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*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)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)2/f*(2*sqrt(a*tan((f*x+exp(1))/2)^2+a)*(-1/2*B*d^2/a^2/sign(tan((f*x+exp(1))/2)+1)+1
/2*B*d^2*tan((f*x+exp(1))/2)/a^2/sign(tan((f*x+exp(1))/2)+1))/(a*tan((f*x+exp(1))/2)^2+a)+2*(1/32*(-29*A*c^2*(
-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+19*A*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*
tan((f*x+exp(1))/2)^2+a))^7+5*B*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7-43*B*d^2*
(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+10*A*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a
*tan((f*x+exp(1))/2)^2+a))^7+38*B*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+75*A*sq
rt(a)*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-133*A*sqrt(a)*d^2*(-sqrt(a)*tan((f*
x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+29*B*sqrt(a)*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x
+exp(1))/2)^2+a))^6+237*B*sqrt(a)*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+58*A*sq
rt(a)*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-266*B*sqrt(a)*c*d*(-sqrt(a)*tan((f*
x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-55*A*a*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1
))/2)^2+a))^5+89*A*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-17*B*a*c^2*(-sqrt(a)
*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-161*B*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((
f*x+exp(1))/2)^2+a))^5-34*A*a*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+178*B*a*c*d
*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-91*A*sqrt(a)*a*c^2*(-sqrt(a)*tan((f*x+exp(1)
)/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+117*A*sqrt(a)*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(
1))/2)^2+a))^4-13*B*sqrt(a)*a*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-221*B*sqrt(
a)*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-26*A*sqrt(a)*a*c*d*(-sqrt(a)*tan((f*
x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+234*B*sqrt(a)*a*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((
f*x+exp(1))/2)^2+a))^4+A*a^2*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+17*A*a^2*d^2
*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-9*B*a^2*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sq
rt(a*tan((f*x+exp(1))/2)^2+a))^3-25*B*a^2*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3
-18*A*a^2*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+34*B*a^2*c*d*(-sqrt(a)*tan((f*x
+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+27*A*a^3*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(
1))/2)^2+a))-53*A*a^3*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+65*A*sqrt(a)*a^2*c^2*
(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-47*A*sqrt(a)*a^2*d^2*(-sqrt(a)*tan((f*x+exp(1
))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+13*B*a^3*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)
^2+a))+93*B*a^3*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-9*B*sqrt(a)*a^2*c^2*(-sqrt(
a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+103*B*sqrt(a)*a^2*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+
sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+26*A*a^3*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))
-18*A*sqrt(a)*a^2*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-106*B*a^3*c*d*(-sqrt(a)
*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-94*B*sqrt(a)*a^2*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(
a*tan((f*x+exp(1))/2)^2+a))^2+7*A*sqrt(a)*a^3*c^2-9*A*sqrt(a)*a^3*d^2+B*sqrt(a)*a^3*c^2+17*B*sqrt(a)*a^3*d^2+2
*A*sqrt(a)*a^3*c*d-18*B*sqrt(a)*a^3*c*d)/a^2/(-(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^
2+2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+a)^4/sign(tan((f*x+exp(1))/2)+1)+1/
32*(3*A*c^2+19*A*d^2+5*B*c^2-75*B*d^2+10*A*c*d+38*B*c*d)*atan((-sqrt(a)*tan((f*x+exp(1))/2)-sqrt(a)+sqrt(a*tan
((f*x+exp(1))/2)^2+a))/sqrt(2)/sqrt(-a))/sqrt(2)/a^2/sqrt(-a)/sign(tan((f*x+exp(1))/2)+1)))
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maple [B] time = 2.04, size = 982, normalized size = 4.48 \[ \text {result too large to display} \]
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))^(5/2),x)
[Out]
-1/32*(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+10*A*2^(1/2)*arct
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+19*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+5*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+38*B*2^(1/2)*arctan
h(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-75*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+64*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^2-10*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-19*A*2^(1/2)*ar
ctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-64*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)-5*B*2^(1/2)*
arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-38*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+75*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)*cos(f*x+e)^2-
6*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2-20*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d+26*A*(a-a*sin(f*x+e))^(3/2)*a^(
1/2)*d^2+20*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2+24*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d-44*A*d^2*a^(3/2)*(a-a
*sin(f*x+e))^(1/2)+6*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+20*A*2^(1/2)*arctan
h(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+38*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/
a^(1/2))*a^2*d^2-10*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2+52*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d-42*B*d^2*(a-a
*sin(f*x+e))^(3/2)*a^(1/2)+12*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2-88*B*c*d*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+204
*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+10*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
+76*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-150*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)*(-a*(sin(f*x+e)-1))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
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))^(5/2),x, algorithm="maxima")
[Out]
integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(5/2),x)
[Out]
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(5/2), x)
________________________________________________________________________________________
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+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(5/2),x)
[Out]
Timed out
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