3.305 \(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=265 \[ \frac {a^{5/2} (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{7/2} f (c+d)^{3/2}}-\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 f (c+d) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d^2 f (c+d)}+\frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))} \]

[Out]

a^(5/2)*(c-d)*(A*d*(3*c+5*d)-B*(5*c^2+5*c*d-2*d^2))*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*
x+e))^(1/2))/d^(7/2)/(c+d)^(3/2)/f+a*(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/d/(c+d)/f/(c+d*sin(f*x+e))-1
/3*a^3*(3*A*d*(3*c+d)-B*(15*c^2-5*c*d-14*d^2))*cos(f*x+e)/d^3/(c+d)/f/(a+a*sin(f*x+e))^(1/2)-1/3*a^2*(-3*A*d+5
*B*c+2*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d^2/(c+d)/f

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Rubi [A]  time = 0.94, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2975, 2976, 2981, 2773, 208} \[ -\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 f (c+d) \sqrt {a \sin (e+f x)+a}}+\frac {a^{5/2} (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{7/2} f (c+d)^{3/2}}-\frac {a^2 (-3 A d+5 B c+2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d^2 f (c+d)}+\frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{d f (c+d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

(a^(5/2)*(c - d)*(A*d*(3*c + 5*d) - B*(5*c^2 + 5*c*d - 2*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c
+ d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(7/2)*(c + d)^(3/2)*f) - (a^3*(3*A*d*(3*c + d) - B*(15*c^2 - 5*c*d - 14*d^
2))*Cos[e + f*x])/(3*d^3*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*(5*B*c - 3*A*d + 2*B*d)*Cos[e + f*x]*Sqrt[
a + a*Sin[e + f*x]])/(3*d^2*(c + d)*f) + (a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(d*(c + d)*f*
(c + d*Sin[e + f*x]))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx &=\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))}+\frac {\int \frac {(a+a \sin (e+f x))^{3/2} \left (-\frac {1}{2} a (3 B c-5 A d-2 B d)+\frac {1}{2} a (5 B c-3 A d+2 B d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=-\frac {a^2 (5 B c-3 A d+2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))}+\frac {2 \int \frac {\sqrt {a+a \sin (e+f x)} \left (-\frac {1}{4} a^2 \left (3 A (c-5 d) d-B \left (5 c^2-7 c d+6 d^2\right )\right )+\frac {1}{4} a^2 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{3 d^2 (c+d)}\\ &=-\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 (c+d) f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 (5 B c-3 A d+2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))}-\frac {\left (a^2 (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d^3 (c+d)}\\ &=-\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 (c+d) f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 (5 B c-3 A d+2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))}+\frac {\left (a^3 (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{d^3 (c+d) f}\\ &=\frac {a^{5/2} (c-d) \left (A d (3 c+5 d)-B \left (5 c^2+5 c d-2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{7/2} (c+d)^{3/2} f}-\frac {a^3 \left (3 A d (3 c+d)-B \left (15 c^2-5 c d-14 d^2\right )\right ) \cos (e+f x)}{3 d^3 (c+d) f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 (5 B c-3 A d+2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{d (c+d) f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 5.97, size = 460, normalized size = 1.74 \[ \frac {(a (\sin (e+f x)+1))^{5/2} \left (\frac {3 (c-d) \left (B \left (5 c^2+5 c d-2 d^2\right )-A d (3 c+5 d)\right ) \left (2 \log \left (\sqrt {d} \sqrt {c+d} \left (\tan ^2\left (\frac {1}{4} (e+f x)\right )+2 \tan \left (\frac {1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}-\frac {3 (c-d) \left (B \left (5 c^2+5 c d-2 d^2\right )-A d (3 c+5 d)\right ) \left (2 \log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}+12 \sqrt {d} (2 A d-4 B c+5 B d) \sin \left (\frac {1}{2} (e+f x)\right )-12 \sqrt {d} (2 A d-4 B c+5 B d) \cos \left (\frac {1}{2} (e+f x)\right )-\frac {12 \sqrt {d} (c-d)^2 (A d-B c) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}-4 B d^{3/2} \sin \left (\frac {3}{2} (e+f x)\right )-4 B d^{3/2} \cos \left (\frac {3}{2} (e+f x)\right )\right )}{12 d^{7/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(-12*Sqrt[d]*(-4*B*c + 2*A*d + 5*B*d)*Cos[(e + f*x)/2] - 4*B*d^(3/2)*Cos[(3*(e +
f*x))/2] - (3*(c - d)*(-(A*d*(3*c + 5*d)) + B*(5*c^2 + 5*c*d - 2*d^2))*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] +
2*Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos[(e + f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/
2]))]))/(c + d)^(3/2) + (3*(c - d)*(-(A*d*(3*c + 5*d)) + B*(5*c^2 + 5*c*d - 2*d^2))*(e + f*x - 2*Log[Sec[(e +
f*x)/4]^2] + 2*Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]*(-1 + 2*Tan[(e + f*x)/4] + Tan[(e + f*x)/4
]^2)]))/(c + d)^(3/2) + 12*Sqrt[d]*(-4*B*c + 2*A*d + 5*B*d)*Sin[(e + f*x)/2] - (12*(c - d)^2*Sqrt[d]*(-(B*c) +
A*d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x])) - 4*B*d^(3/2)*Sin[(3*(e + f*x))/2]
))/(12*d^(7/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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fricas [B]  time = 2.59, size = 2046, normalized size = 7.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/12*(3*(5*B*a^2*c^4 - (3*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^2*c^2*d^2 + (3*A - 5*B)*a^2*c*d^3 + (5*A + 2*B)
*a^2*d^4 - (5*B*a^2*c^3*d - 3*A*a^2*c^2*d^2 - (2*A + 7*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4)*cos(f*x + e)^2 + (5
*B*a^2*c^4 - 3*A*a^2*c^3*d - (2*A + 7*B)*a^2*c^2*d^2 + (5*A + 2*B)*a^2*c*d^3)*cos(f*x + e) + (5*B*a^2*c^4 - (3
*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^2*c^2*d^2 + (3*A - 5*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4 + (5*B*a^2*c^3*d -
3*A*a^2*c^2*d^2 - (2*A + 7*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a/(c*d + d^2)
)*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^
2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 +
(c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9
*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*s
in(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e)
+ (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(15*B*a^2*c^3 - (9*A + 20*
B)*a^2*c^2*d + 3*(2*A - 3*B)*a^2*c*d^2 + (3*A + 14*B)*a^2*d^3 + 2*(B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e)^3 + 2
*(5*B*a^2*c^2*d - (3*A + 2*B)*a^2*c*d^2 - (3*A + 7*B)*a^2*d^3)*cos(f*x + e)^2 + (15*B*a^2*c^3 - (9*A + 10*B)*a
^2*c^2*d - 15*B*a^2*c*d^2 - (3*A + 2*B)*a^2*d^3)*cos(f*x + e) - (15*B*a^2*c^3 - (9*A + 20*B)*a^2*c^2*d + 3*(2*
A - 3*B)*a^2*c*d^2 + (3*A + 14*B)*a^2*d^3 + 2*(B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e)^2 - 2*(5*B*a^2*c^2*d - 3*
(A + B)*a^2*c*d^2 - (3*A + 8*B)*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^4 + d^5)*
f*cos(f*x + e)^2 - (c^2*d^3 + c*d^4)*f*cos(f*x + e) - (c^2*d^3 + 2*c*d^4 + d^5)*f - ((c*d^4 + d^5)*f*cos(f*x +
e) + (c^2*d^3 + 2*c*d^4 + d^5)*f)*sin(f*x + e)), 1/6*(3*(5*B*a^2*c^4 - (3*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^
2*c^2*d^2 + (3*A - 5*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4 - (5*B*a^2*c^3*d - 3*A*a^2*c^2*d^2 - (2*A + 7*B)*a^2*c
*d^3 + (5*A + 2*B)*a^2*d^4)*cos(f*x + e)^2 + (5*B*a^2*c^4 - 3*A*a^2*c^3*d - (2*A + 7*B)*a^2*c^2*d^2 + (5*A + 2
*B)*a^2*c*d^3)*cos(f*x + e) + (5*B*a^2*c^4 - (3*A - 5*B)*a^2*c^3*d - (5*A + 7*B)*a^2*c^2*d^2 + (3*A - 5*B)*a^2
*c*d^3 + (5*A + 2*B)*a^2*d^4 + (5*B*a^2*c^3*d - 3*A*a^2*c^2*d^2 - (2*A + 7*B)*a^2*c*d^3 + (5*A + 2*B)*a^2*d^4)
*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*
d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(15*B*a^2*c^3 - (9*A + 20*B)*a^2*c^2*d + 3*(2*A - 3*B)*a^2*c*d^2
+ (3*A + 14*B)*a^2*d^3 + 2*(B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e)^3 + 2*(5*B*a^2*c^2*d - (3*A + 2*B)*a^2*c*d^
2 - (3*A + 7*B)*a^2*d^3)*cos(f*x + e)^2 + (15*B*a^2*c^3 - (9*A + 10*B)*a^2*c^2*d - 15*B*a^2*c*d^2 - (3*A + 2*B
)*a^2*d^3)*cos(f*x + e) - (15*B*a^2*c^3 - (9*A + 20*B)*a^2*c^2*d + 3*(2*A - 3*B)*a^2*c*d^2 + (3*A + 14*B)*a^2*
d^3 + 2*(B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e)^2 - 2*(5*B*a^2*c^2*d - 3*(A + B)*a^2*c*d^2 - (3*A + 8*B)*a^2*d^
3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^4 + d^5)*f*cos(f*x + e)^2 - (c^2*d^3 + c*d^4)*f
*cos(f*x + e) - (c^2*d^3 + 2*c*d^4 + d^5)*f - ((c*d^4 + d^5)*f*cos(f*x + e) + (c^2*d^3 + 2*c*d^4 + d^5)*f)*sin
(f*x + e))]

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giac [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+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 2.54, size = 932, normalized size = 3.52 \[ \frac {a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sin \left (f x +e \right ) d \left (9 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{2} d +6 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c \,d^{2}-15 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} d^{3}-15 a^{2} \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) B \,c^{3}+21 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c \,d^{2}-6 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} d^{3}+2 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c d +2 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{2}-6 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a c d -6 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{2}+12 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2}-6 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a c d -18 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{2}\right )+9 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{3} d +6 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{2} d^{2}-15 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c \,d^{3}+2 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c^{2} d +2 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c \,d^{2}-15 a^{2} \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) B \,c^{4}+21 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{2} d^{2}-6 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c \,d^{3}-9 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2} d -3 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{3}+15 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{3}-12 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2} d -15 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a c \,d^{2}\right )}{3 d^{3} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(sin(f*x+e)*d*(9*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^
2)^(1/2))*a^2*c^2*d+6*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^2-15*A*arctanh((a-a*sin(
f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3-15*a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*B*c^
3+21*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^2-6*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a
*c*d+a*d^2)^(1/2))*a^2*d^3+2*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d+2*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d
)*d)^(1/2)*d^2-6*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-6*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)
*a*d^2+12*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2-6*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-
18*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2)+9*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))
*a^2*c^3*d+6*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2-15*A*arctanh((a-a*sin(f*x+e))
^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3+2*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d+2*B*(a-a*sin(f*x+e)
)^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2-15*a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*B*c^4+21*B*arctan
h((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2-6*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)
^(1/2))*a^2*c*d^3-9*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2*d-3*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^
(1/2)*a*d^3+15*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^3-12*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*
a*c^2*d-15*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d^2)/d^3/(c+d)/(c+d*sin(f*x+e))/(a*(c+d)*d)^(1/2)/co
s(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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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