3.6.42 \(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx\) [542]

Optimal. Leaf size=194 \[ -\frac {a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \]

[Out]

-1/4*a^(5/2)*(3*c^2+10*c*d+19*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^(5
/2)/(c+d)^(5/2)/f+3/4*a^3*(c-d)*(c+3*d)*cos(f*x+e)/d^2/(c+d)^2/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)+1/2*a
^2*(c-d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/(c+d)/f/(c+d*sin(f*x+e))^2

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Rubi [A]
time = 0.30, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2841, 3059, 2852, 214} \begin {gather*} -\frac {a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{4 d^{5/2} f (c+d)^{5/2}}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 f (c+d)^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

-1/4*(a^(5/2)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e +
f*x]])])/(d^(5/2)*(c + d)^(5/2)*f) + (a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*d*(c + d)*f*(c +
d*Sin[e + f*x])^2) + (3*a^3*(c - d)*(c + 3*d)*Cos[e + f*x])/(4*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d
*Sin[e + f*x]))

Rule 214

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

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2852

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 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx &=\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (c-9 d)-\frac {1}{2} a (3 c+5 d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {\left (a^2 \left (3 c^2+10 c d+19 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d^2 (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {\left (a^3 \left (3 c^2+10 c d+19 d^2\right )\right ) \text {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 )}{4 d^2 (c+d)^2 f}\\ &=-\frac {a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 3.20, size = 379, normalized size = 1.95 \begin {gather*} \frac {(a (1+\sin (e+f x)))^{5/2} \left (-\frac {\left (3 c^2+10 c d+19 d^2\right ) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (c+d+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{(c+d)^{5/2}}+\frac {\left (3 c^2+10 c d+19 d^2\right ) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left ((c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \left (-1+2 \tan \left (\frac {1}{4} (e+f x)\right )+\tan ^2\left (\frac {1}{4} (e+f x)\right )\right )\right )\right )}{(c+d)^{5/2}}-\frac {8 (c-d)^2 \sqrt {d} \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))^2}-\frac {4 \sqrt {d} \left (-5 c^2-6 c d+11 d^2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^2 (c+d \sin (e+f x))}\right )}{16 d^{5/2} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(-(((3*c^2 + 10*c*d + 19*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)^(5/2)) + ((3*c^2 + 10*c*d + 19*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)^(5/2) - (8*(c - d)^2*Sqrt[d]
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x])^2) - (4*Sqrt[d]*(-5*c^2 - 6*c*d + 11*d^2
)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)^2*(c + d*Sin[e + f*x]))))/(16*d^(5/2)*f*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(566\) vs. \(2(170)=340\).
time = 5.64, size = 567, normalized size = 2.92

method	result	size
default	\(-\frac {a \left (2 \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} c d \left (3 c^{2}+10 c d +19 d^{2}\right )-\arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} d^{2} \left (3 c^{2}+10 c d +19 d^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} c^{4}+10 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} c^{3} d +22 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} c^{2} d^{2}+10 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} c \,d^{3}+19 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{2} d^{4}+5 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c^{2} d +6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c \,d^{2}-11 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{3}-3 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{3}-13 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2} d +3 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a c \,d^{2}+13 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{3}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 \sqrt {a \left (c +d \right ) d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(567\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*a*(2*sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(3*c^2+10*c*d+19*d^2)-arcta
nh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(3*c^2+10*c*d+19*d^2)*cos(f*x+e)^2+3*arctanh((a-a*sin
(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^4+10*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^3
*d+22*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2+10*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a
*c*d+a*d^2)^(1/2))*a^2*c*d^3+19*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^4+5*(a-a*sin(f*x+e
))^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d+6*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2-11*(a-a*sin(f*x+e))^(3/2)*(a
*(c+d)*d)^(1/2)*d^3-3*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^3-13*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/
2)*a*c^2*d+3*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d^2+13*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^
3)*(-a*(sin(f*x+e)-1))^(1/2)*(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/d^2/cos(f*x+e)/(a+a*s
in(f*x+e))^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 859 vs. \(2 (178) = 356\).
time = 0.56, size = 2048, normalized size = 10.56 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*((3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3
+ 19*a^2*d^4)*cos(f*x + e)^3 - (6*a^2*c^3*d + 23*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + (3
*a^2*c^4 + 10*a^2*c^3*d + 22*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 + 16*a^2*c^3*d
+ 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + 2
*(3*a^2*c^3*d + 10*a^2*c^2*d^2 + 19*a^2*c*d^3)*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)*co
s(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))*sin(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*(3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9
*a^2*d^3 + (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 8*a^2*c^2*d - 9*a^2*c*d^2 -
2*a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 - (5*a^2*c^2*d + 6*a^2*c*d^2 - 1
1*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 +
(2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*
f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x +
e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*s
in(f*x + e)), 1/8*((3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 1
0*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^3 - (6*a^2*c^3*d + 23*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x
+ e)^2 + (3*a^2*c^4 + 10*a^2*c^3*d + 22*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 + 1
6*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x
+ e)^2 + 2*(3*a^2*c^3*d + 10*a^2*c^2*d^2 + 19*a^2*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arc
tan(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*(3*a^2*
c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 + (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e)^2 + (3*a^
2*c^3 + 8*a^2*c^2*d - 9*a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*
d^3 - (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4
+ 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c
^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c
^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d
^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (178) = 356\).
time = 0.56, size = 448, normalized size = 2.31 \begin {gather*} -\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (3 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 10 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 19 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} \sqrt {-c d - d^{2}}} - \frac {2 \, {\left (10 \, a^{2} c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 12 \, a^{2} c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 22 \, a^{2} d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 13 \, a^{2} c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a^{2} c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, a^{2} d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} {\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}\right )}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(3*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 10*a^2*c*d*sgn(cos(-1/4*pi + 1/
2*f*x + 1/2*e)) + 19*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2
*e)/sqrt(-c*d - d^2))/((c^2*d^2 + 2*c*d^3 + d^4)*sqrt(-c*d - d^2)) - 2*(10*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 12*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1
/2*f*x + 1/2*e)^3 - 22*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 3*a^2*c^
3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 13*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 3*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f
*x + 1/2*e) + 13*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((c^2*d^2 + 2*c*d
^3 + d^4)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)^2))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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