3.6.35 \(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx\) [535]

Optimal. Leaf size=179 \[ -\frac {a^{3/2} (c+7 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{3/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \]

[Out]

-1/4*a^(3/2)*(c+7*d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(3/2)/(c+d)^(5/2
)/f+1/2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2)-1/4*a^2*(c+7*d)*cos(f*x+e)/d/
(c+d)^2/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2841, 21, 2851, 2852, 214} \begin {gather*} -\frac {a^{3/2} (c+7 d) \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^{3/2} f (c+d)^{5/2}}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d 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)}{2 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a^(3/2)*(c + 7*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(3/
2)*(c + d)^(5/2)*f) + (a^2*(c - d)*Cos[e + f*x])/(2*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^
2) - (a^2*(c + 7*d)*Cos[e + f*x])/(4*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 2851

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

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]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a \int \frac {-\frac {1}{2} a (c+7 d)-\frac {1}{2} a (c+7 d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {(a (c+7 d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx}{4 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {(a (c+7 d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {\left (a^2 (c+7 d)\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 (c+d)^2 f}\\ &=-\frac {a^{3/2} (c+7 d) \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^{3/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (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 = 2.72, size = 313, normalized size = 1.75 \begin {gather*} \frac {(a (1+\sin (e+f x)))^{3/2} \left (-\frac {2 (c+7 d) \left (\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 )-\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 {4 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right ) \left (-c^2+7 c d+2 d^2+d (c+7 d) \sin (e+f x)\right )}{(c+d)^2 (c+d \sin (e+f x))^2}+\frac {4 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right ) \left (-c^2+7 c d+2 d^2+d (c+7 d) \sin (e+f x)\right )}{(c+d)^2 (c+d \sin (e+f x))^2}\right )}{16 d^{3/2} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*((-2*(c + 7*d)*(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]))] - 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) - (4*Sqrt[d]*Cos[(e + f*x)/2]*(-c^2 + 7*c*d + 2*d^2
+ d*(c + 7*d)*Sin[e + f*x]))/((c + d)^2*(c + d*Sin[e + f*x])^2) + (4*Sqrt[d]*Sin[(e + f*x)/2]*(-c^2 + 7*c*d +
2*d^2 + d*(c + 7*d)*Sin[e + f*x]))/((c + d)^2*(c + d*Sin[e + f*x])^2)))/(16*d^(3/2)*f*(Cos[(e + f*x)/2] + Sin[
(e + f*x)/2])^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(428\) vs. \(2(155)=310\).
time = 5.72, size = 429, normalized size = 2.40

method	result	size
default	\(\frac {\left (-\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) a^{2} c \,d^{2}-7 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) a^{2} d^{3}-2 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c^{2} d -14 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c \,d^{2}+\left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c d +7 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{2}-\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{3}-7 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{2} d +\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2}-8 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a c d -9 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 \sqrt {a \left (c +d \right ) d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*c*d^2-7*arctanh((-a*(sin(f*x+e)-
1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*d^3-2*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*s
in(f*x+e)*a^2*c^2*d-14*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a^2*c*d^2+(-a*(sin(f*
x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*c*d+7*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*d^2-arctanh((-a*(sin(f*x+e)
-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^3-7*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^2*d+(-a
*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2-8*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-9*(-a*(sin(
f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2)*(-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/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 \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))^(3/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)/(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. 641 vs. \(2 (163) = 326\).
time = 0.52, size = 1612, normalized size = 9.01 \begin {gather*} \left [-\frac {{\left (a c^{3} + 9 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3} - {\left (a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (2 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{3} + 7 \, a c^{2} d + a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right ) + {\left (a c^{3} + 9 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3} - {\left (a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a c^{2} d + 7 \, a c d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {a}{c d + d^{2}}} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} - {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} + {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {a}{c d + d^{2}}} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (a c^{2} - 6 \, a c d + 5 \, a d^{2} - {\left (a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{2} - 7 \, a c d - 2 \, a d^{2}\right )} \cos \left (f x + e\right ) - {\left (a c^{2} - 6 \, a c d + 5 \, a d^{2} + {\left (a c d + 7 \, a 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}}{16 \, {\left ({\left (c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{3} + {\left (2 \, c^{3} d^{2} + 5 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{4} d + 2 \, c^{3} d^{2} + 2 \, c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} d + 4 \, c^{3} d^{2} + 6 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f + {\left ({\left (c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d^{2} + 2 \, c^{2} d^{3} + c d^{4}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} d + 4 \, c^{3} d^{2} + 6 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f\right )} \sin \left (f x + e\right )\right )}}, \frac {{\left (a c^{3} + 9 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3} - {\left (a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (2 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{3} + 7 \, a c^{2} d + a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right ) + {\left (a c^{3} + 9 \, a c^{2} d + 15 \, a c d^{2} + 7 \, a d^{3} - {\left (a c d^{2} + 7 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a c^{2} d + 7 \, a c d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {a}{c d + d^{2}}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {a}{c d + d^{2}}}}{2 \, a \cos \left (f x + e\right )}\right ) - 2 \, {\left (a c^{2} - 6 \, a c d + 5 \, a d^{2} - {\left (a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{2} - 7 \, a c d - 2 \, a d^{2}\right )} \cos \left (f x + e\right ) - {\left (a c^{2} - 6 \, a c d + 5 \, a d^{2} + {\left (a c d + 7 \, a 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}}{8 \, {\left ({\left (c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{3} + {\left (2 \, c^{3} d^{2} + 5 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{4} d + 2 \, c^{3} d^{2} + 2 \, c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} d + 4 \, c^{3} d^{2} + 6 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f + {\left ({\left (c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d^{2} + 2 \, c^{2} d^{3} + c d^{4}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} d + 4 \, c^{3} d^{2} + 6 \, c^{2} d^{3} + 4 \, c d^{4} + d^{5}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^3 - (2*a*c^2*d + 15*a*c*d
^2 + 7*a*d^3)*cos(f*x + e)^2 + (a*c^3 + 7*a*c^2*d + a*c*d^2 + 7*a*d^3)*cos(f*x + e) + (a*c^3 + 9*a*c^2*d + 15*
a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + 2*(a*c^2*d + 7*a*c*d^2)*cos(f*x + e))*sin(f*x + e))*s
qrt(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))*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*(a*c^2 -
6*a*c*d + 5*a*d^2 - (a*c*d + 7*a*d^2)*cos(f*x + e)^2 + (a*c^2 - 7*a*c*d - 2*a*d^2)*cos(f*x + e) - (a*c^2 - 6*a
*c*d + 5*a*d^2 + (a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^3 + 2*c*d^4 +
d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2
*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4
+ d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^4)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4
*c*d^4 + d^5)*f)*sin(f*x + e)), 1/8*((a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x +
e)^3 - (2*a*c^2*d + 15*a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + (a*c^3 + 7*a*c^2*d + a*c*d^2 + 7*a*d^3)*cos(f*x +
e) + (a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + 2*(a*c^2*d + 7*a*c*d^2)*
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*(a*c^2 - 6*a*c*d + 5*a*d^2 - (a*c*d + 7*a*d^2)*cos(f*x + e)^2 + (
a*c^2 - 7*a*c*d - 2*a*d^2)*cos(f*x + e) - (a*c^2 - 6*a*c*d + 5*a*d^2 + (a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x
+ e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^
4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^
2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^
4)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*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))**(3/2)/(c+d*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]
time = 0.59, size = 325, normalized size = 1.82 \begin {gather*} -\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, a d \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 \, c d^{2} + d^{3}\right )} \sqrt {-c d - d^{2}}} + \frac {2 \, {\left (2 \, a c 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} + 14 \, a 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} + a c^{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 ) - 8 \, a c 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 ) - 9 \, a 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 )\right )}}{{\left (c^{2} d + 2 \, c d^{2} + d^{3}\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))^(3/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 7*a*d*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*c*d^2 + d^3)*sqrt(-c*d - d
^2)) + 2*(2*a*c*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 + 14*a*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 + a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*
pi + 1/2*f*x + 1/2*e) - 8*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 9*a*d^2*s
gn(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*c*d^2 + d^3)*(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 )}^{3/2}}{{\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))^(3/2)/(c + d*sin(e + f*x))^3,x)

[Out]

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

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