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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 \[ -\frac {(c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (9 A c+39 A d+15 B c-95 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{48 a^3 f}+\frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {(3 A c+9 A d+5 B c-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-1/16*(3*A*c+9*A*d+5*B*c-17*B*d)*cos(f*x+e)*(c+d*sin(f*x+e))^2/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(A-B)*cos(f*x+e)
*(c+d*sin(f*x+e))^3/f/(a+a*sin(f*x+e))^(5/2)-1/32*(c-d)*(B*(5*c^2+62*c*d-163*d^2)+3*A*(c^2+6*c*d+25*d^2))*arct
anh(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)+1/24*d*(A*(9*c^2+36*c*d-93*d^2)+B
*(15*c^2-228*c*d+197*d^2))*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)+1/48*d^2*(9*A*c+39*A*d+15*B*c-95*B*d)*cos(f
*x+e)*(a+a*sin(f*x+e))^(1/2)/a^3/f

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Rubi [A]  time = 1.06, antiderivative size = 308, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [C]  time = 1.79, size = 523, normalized size = 1.70 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left ((3+3 i) (-1)^{3/4} (c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+(24+24 i) d^2 (-2 A d-6 B c+5 B d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4+(24+24 i) d^2 (2 A d+6 B c-5 B d) \left (\sin \left (\frac {1}{2} (e+f x)\right )+i \cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (A-B) (c-d)^3 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 (3 A (c+7 d)+B (5 c-29 d)) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3+6 (c-d)^2 (3 A (c+7 d)+B (5 c-29 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-12 (A-B) (c-d)^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-16 B d^3 \cos \left (\frac {3}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4-16 B d^3 \sin \left (\frac {3}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{48 f (a (\sin (e+f x)+1))^{5/2}} \]

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))

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fricas [B]  time = 0.50, size = 980, normalized size = 3.18 \[ \text {result too large to display} \]

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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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))^3/(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)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 sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (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 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)/(a*tan((f*x+exp(1))/2)^2+a)*(tan((f*x+exp(1))
/2)*(tan((f*x+exp(1))/2)*(-1/144*tan((f*x+exp(1))/2)*(-72*A*a^3*d^3*sign(tan((f*x+exp(1))/2)+1)+168*B*a^3*d^3*
sign(tan((f*x+exp(1))/2)+1)-216*B*a^3*c*d^2*sign(tan((f*x+exp(1))/2)+1))/a^4-1/144*(72*A*a^3*d^3*sign(tan((f*x
+exp(1))/2)+1)-216*B*a^3*d^3*sign(tan((f*x+exp(1))/2)+1)+216*B*a^3*c*d^2*sign(tan((f*x+exp(1))/2)+1))/a^4)-1/1
44*(-72*A*a^3*d^3*sign(tan((f*x+exp(1))/2)+1)+216*B*a^3*d^3*sign(tan((f*x+exp(1))/2)+1)-216*B*a^3*c*d^2*sign(t
an((f*x+exp(1))/2)+1))/a^4)-1/144*(72*A*a^3*d^3*sign(tan((f*x+exp(1))/2)+1)-168*B*a^3*d^3*sign(tan((f*x+exp(1)
)/2)+1)+216*B*a^3*c*d^2*sign(tan((f*x+exp(1))/2)+1))/a^4)+2*(1/32*(-29*A*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqr
t(a*tan((f*x+exp(1))/2)^2+a))^7-43*A*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+5*B*
c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+67*B*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sq
rt(a*tan((f*x+exp(1))/2)^2+a))^7+57*A*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+1
5*A*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7-129*B*c*d^2*(-sqrt(a)*tan((f*x+exp(
1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+57*B*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^
2+a))^7+75*A*sqrt(a)*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+237*A*sqrt(a)*d^3*(-
sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+29*B*sqrt(a)*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+
sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-341*B*sqrt(a)*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^
2+a))^6-399*A*sqrt(a)*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+87*A*sqrt(a)*c^2*
d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+711*B*sqrt(a)*c*d^2*(-sqrt(a)*tan((f*x+exp(
1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-399*B*sqrt(a)*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+ex
p(1))/2)^2+a))^6-55*A*a*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-161*A*a*d^3*(-sqr
t(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-17*B*a*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*ta
n((f*x+exp(1))/2)^2+a))^5+233*B*a*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+267*A*a
*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-51*A*a*c^2*d*(-sqrt(a)*tan((f*x+exp(1)
)/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-483*B*a*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)
^2+a))^5+267*B*a*c^2*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^3*(-s
qrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-221*A*sqrt(a)*a*d^3*(-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^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/
2)^2+a))^4+325*B*sqrt(a)*a*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+351*A*sqrt(a)*
a*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-39*A*sqrt(a)*a*c^2*d*(-sqrt(a)*tan((f
*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-663*B*sqrt(a)*a*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*ta
n((f*x+exp(1))/2)^2+a))^4+351*B*sqrt(a)*a*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))
^4+A*a^2*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-25*A*a^2*d^3*(-sqrt(a)*tan((f*x+
exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-9*B*a^2*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1)
)/2)^2+a))^3+33*B*a^2*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+51*A*a^2*c*d^2*(-sq
rt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-27*A*a^2*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt
(a*tan((f*x+exp(1))/2)^2+a))^3-75*B*a^2*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3
+51*B*a^2*c^2*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^3*(-sqrt(a)*tan((f
*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+93*A*a^3*d^3*(-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^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+103*A*sqrt(
a)*a^2*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+13*B*a^3*c^3*(-sqrt(a)*tan((f*x+ex
p(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-133*B*a^3*d^3*(-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^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-159*B*sqrt(a)*a^
2*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-159*A*a^3*c*d^2*(-sqrt(a)*tan((f*x+exp(
1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+39*A*a^3*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2
)^2+a))-141*A*sqrt(a)*a^2*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-27*A*sqrt(a)*
a^2*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+279*B*a^3*c*d^2*(-sqrt(a)*tan((f*x+
exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-159*B*a^3*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(
1))/2)^2+a))+309*B*sqrt(a)*a^2*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-141*B*sq
rt(a)*a^2*c^2*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^3+17*A*sqrt
(a)*a^3*d^3+B*sqrt(a)*a^3*c^3-25*B*sqrt(a)*a^3*d^3-27*A*sqrt(a)*a^3*c*d^2+3*A*sqrt(a)*a^3*c^2*d+51*B*sqrt(a)*a
^3*c*d^2-27*B*sqrt(a)*a^3*c^2*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^3-75*A*d^3+5*B*c^3+163*B*d^3+57*A*c*d^2+15*A*c^2*d-225*B*c*d^2+57*B*c^2*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 = 2.47, size = 1438, normalized size = 4.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^(5/2),x)

[Out]

1/96/a^(9/2)*((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)*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)*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)*cos(f*x+e)^2-2*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)*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)*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)-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+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^
3-60*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^3-612*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^3-342*A*2^(1/2)*a
rctanh(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^2-18*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+396*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2*d-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^2-234*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+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)*(-a*(sin(f*x+e)-1))^(1/2)/(
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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{3}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^3)/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^3)/(a + a*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+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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