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

Optimal. Leaf size=166 \[ \frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt {a \sin (e+f x)+a}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 166, 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 = {2762, 2981, 2773, 208} \[ -\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt {a \sin (e+f x)+a}}+\frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(c - d)*(3*c + 5*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(
d^(5/2)*(c + d)^(3/2)*f) - (a^3*(3*c + d)*Cos[e + f*x])/(d^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]) + (a^2*(c - d
)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(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 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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 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 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}}{(c+d \sin (e+f x))^2} \, dx &=\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac {a \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (c-5 d)-\frac {1}{2} a (3 c+d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac {\left (a^2 (c-d) (3 c+5 d)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}+\frac {\left (a^3 (c-d) (3 c+5 d)\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^2 (c+d) f}\\ &=\frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{5/2} (c+d)^{3/2} f}-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [B]  time = 4.09, size = 350, normalized size = 2.11 \[ \frac {(a (\sin (e+f x)+1))^{5/2} \left (\frac {\left (-3 c^2-2 c d+5 d^2\right ) \left (2 \log \left (\sqrt {d} \sqrt {c+d} \left (\tan ^2\left (\frac {1}{4} (e+f x)\right )+2 \tan \left (\frac {1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}+\frac {\left (3 c^2+2 c d-5 d^2\right ) \left (2 \log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}-\frac {4 \sqrt {d} (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+8 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-8 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )}{4 d^{5/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(-8*Sqrt[d]*Cos[(e + f*x)/2] + ((3*c^2 + 2*c*d - 5*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^2 - 2*c*d + 5*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
) + 8*Sqrt[d]*Sin[(e + f*x)/2] - (4*(c - d)^2*Sqrt[d]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*S
in[e + f*x]))))/(4*d^(5/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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fricas [B]  time = 0.63, size = 1322, normalized size = 7.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d + 2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e
)^2 + (3*a^2*c^3 + 2*a^2*c^2*d - 5*a^2*c*d^2)*cos(f*x + e) + (3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^
3 + (3*a^2*c^2*d + 2*a^2*c*d^2 - 5*a^2*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)*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*co
s(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^2 - 2*a^2*c*d - a^2*d^2 + 2*(a^2*c*d
+ a^2*d^2)*cos(f*x + e)^2 + (3*a^2*c^2 + a^2*d^2)*cos(f*x + e) - (3*a^2*c^2 - 2*a^2*c*d - a^2*d^2 - 2*(a^2*c*d
+ a^2*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^2*d^2 +
c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*c*d^3 + d^
4)*f)*sin(f*x + e)), -1/2*((3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d + 2*a^2*c*d^2 - 5
*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 2*a^2*c^2*d - 5*a^2*c*d^2)*cos(f*x + e) + (3*a^2*c^3 + 5*a^2*c^2*d - 3
*a^2*c*d^2 - 5*a^2*d^3 + (3*a^2*c^2*d + 2*a^2*c*d^2 - 5*a^2*d^3)*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*
(3*a^2*c^2 - 2*a^2*c*d - a^2*d^2 + 2*(a^2*c*d + a^2*d^2)*cos(f*x + e)^2 + (3*a^2*c^2 + a^2*d^2)*cos(f*x + e) -
(3*a^2*c^2 - 2*a^2*c*d - a^2*d^2 - 2*(a^2*c*d + a^2*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)
)/((c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 +
d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*c*d^3 + d^4)*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)/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.48, size = 393, normalized size = 2.37 \[ -\frac {a^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-\sin \left (f x +e \right ) d \left (3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2}+2 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c d -5 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,d^{2}-2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c -2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d \right )-3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{3}-2 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2} d +5 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c \,d^{2}+3 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c^{2}+\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d^{2}\right )}{d^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-sin(f*x+e)*d*(3*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)
^(1/2))*a*c^2+2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d-5*arctanh((a-a*sin(f*x+e))^(1/2)*d
/(a*c*d+a*d^2)^(1/2))*a*d^2-2*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c-2*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^
(1/2)*d)-3*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^3-2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c
*d+a*d^2)^(1/2))*a*c^2*d+5*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d^2+3*(a-a*sin(f*x+e))^(1
/2)*(a*(c+d)*d)^(1/2)*c^2+(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*d^2)/d^2/(c+d)/(c+d*sin(f*x+e))/(a*(c+d)*d)
^(1/2)/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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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 = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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