3.570 \(\int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac{16 a \cos (e+f x)}{15 f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{8 a \cos (e+f x)}{15 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{2 a \cos (e+f x)}{5 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}} \]

[Out]

(-2*a*Cos[e + f*x])/(5*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (8*a*Cos[e + f*x])/(15
*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (16*a*Cos[e + f*x])/(15*(c + d)^3*f*Sqrt[a
+ a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.294739, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2772, 2771} \[ -\frac{16 a \cos (e+f x)}{15 f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{8 a \cos (e+f x)}{15 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{2 a \cos (e+f x)}{5 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(-2*a*Cos[e + f*x])/(5*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (8*a*Cos[e + f*x])/(15
*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (16*a*Cos[e + f*x])/(15*(c + d)^3*f*Sqrt[a
+ a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

Rule 2772

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 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{7/2}} \, dx &=-\frac{2 a \cos (e+f x)}{5 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}+\frac{4 \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 (c+d)}\\ &=-\frac{2 a \cos (e+f x)}{5 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{8 a \cos (e+f x)}{15 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{8 \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 (c+d)^2}\\ &=-\frac{2 a \cos (e+f x)}{5 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{8 a \cos (e+f x)}{15 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{16 a \cos (e+f x)}{15 (c+d)^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.393021, size = 128, normalized size = 0.9 \[ -\frac{2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (15 c^2+4 d (5 c+d) \sin (e+f x)+10 c d+8 d^2 \sin ^2(e+f x)+3 d^2\right )}{15 f (c+d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[e + f*x]]/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(15*c^2 + 10*c*d + 3*d^2 + 4*d*(5*c + d)*
Sin[e + f*x] + 8*d^2*Sin[e + f*x]^2))/(15*(c + d)^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c + d*Sin[e + f*x
])^(5/2))

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Maple [B]  time = 0.244, size = 430, normalized size = 3. \begin{align*}{\frac{-4\,c \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{4}-14\,{d}^{5}-30\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{2}{d}^{3}-70\,{c}^{4}d-12\,{c}^{2}{d}^{3}-22\,c{d}^{4}-46\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{5}+44\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{5}+16\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{d}^{5}+30\,{c}^{5}\sin \left ( fx+e \right ) +12\,{c}^{2}{d}^{3}\sin \left ( fx+e \right ) +22\,c{d}^{4}\sin \left ( fx+e \right ) +14\,{d}^{5}\sin \left ( fx+e \right ) -44\,{c}^{3}{d}^{2}+8\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}c{d}^{4}+14\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{3}{d}^{2}+6\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{2}{d}^{3}-30\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}c{d}^{4}+38\,{c}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+26\,c{d}^{4} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{5}-22\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{5}+50\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{4}d+70\,\sin \left ( fx+e \right ){c}^{4}d+44\,\sin \left ( fx+e \right ){c}^{3}{d}^{2}+42\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{c}^{2}{d}^{3}-30\,{c}^{5}}{15\,f \left ( c+d \right ) ^{3}\cos \left ( fx+e \right ) \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+{c}^{2}-{d}^{2} \right ) ^{3}}\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{c+d\sin \left ( fx+e \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(7/2),x)

[Out]

2/15/f/(c+d)^3*(a*(1+sin(f*x+e)))^(1/2)*(c+d*sin(f*x+e))^(1/2)*(-2*c*cos(f*x+e)^4*d^4-7*d^5-15*cos(f*x+e)^2*c^
2*d^3-35*c^4*d-6*c^2*d^3-11*c*d^4-23*cos(f*x+e)^4*d^5+22*cos(f*x+e)^2*d^5+8*cos(f*x+e)^6*d^5+15*c^5*sin(f*x+e)
+6*c^2*d^3*sin(f*x+e)+11*c*d^4*sin(f*x+e)+7*d^5*sin(f*x+e)-22*c^3*d^2+4*sin(f*x+e)*cos(f*x+e)^4*c*d^4+7*sin(f*
x+e)*cos(f*x+e)^2*c^3*d^2+3*sin(f*x+e)*cos(f*x+e)^2*c^2*d^3-15*sin(f*x+e)*cos(f*x+e)^2*c*d^4+19*c^3*cos(f*x+e)
^2*d^2+13*c*d^4*cos(f*x+e)^2+4*sin(f*x+e)*cos(f*x+e)^4*d^5-11*sin(f*x+e)*cos(f*x+e)^2*d^5+25*cos(f*x+e)^2*c^4*
d+35*sin(f*x+e)*c^4*d+22*sin(f*x+e)*c^3*d^2+21*cos(f*x+e)^4*c^2*d^3-15*c^5)/cos(f*x+e)/(cos(f*x+e)^2*d^2+c^2-d
^2)^3

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Maxima [B]  time = 2.27793, size = 734, normalized size = 5.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-2/15*((15*c^3 + 10*c^2*d + 3*c*d^2)*sqrt(a) - (15*c^3 - 60*c^2*d - 25*c*d^2 - 6*d^3)*sqrt(a)*sin(f*x + e)/(co
s(f*x + e) + 1) + (45*c^3 - 40*c^2*d + 93*c*d^2 + 10*d^3)*sqrt(a)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*(9*c
^3 - 22*c^2*d + 13*c*d^2 - 12*d^3)*sqrt(a)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*(9*c^3 - 22*c^2*d + 13*c*d^
2 - 12*d^3)*sqrt(a)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (45*c^3 - 40*c^2*d + 93*c*d^2 + 10*d^3)*sqrt(a)*sin(
f*x + e)^5/(cos(f*x + e) + 1)^5 + (15*c^3 - 60*c^2*d - 25*c*d^2 - 6*d^3)*sqrt(a)*sin(f*x + e)^6/(cos(f*x + e)
+ 1)^6 - (15*c^3 + 10*c^2*d + 3*c*d^2)*sqrt(a)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*(sin(f*x + e)^2/(cos(f*x +
e) + 1)^2 + 1)^3/((c^3 + 3*c^2*d + 3*c*d^2 + d^3 + 3*(c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 3*(c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + (c^3 + 3*c^2*d + 3*c*d^2
+ d^3)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2)^(7/2)*f)

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Fricas [B]  time = 2.60363, size = 1264, normalized size = 8.9 \begin{align*} \frac{2 \,{\left (8 \, d^{2} \cos \left (f x + e\right )^{3} - 4 \,{\left (5 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, c^{2} + 10 \, c d - 7 \, d^{2} -{\left (15 \, c^{2} + 10 \, c d + 11 \, d^{2}\right )} \cos \left (f x + e\right ) -{\left (8 \, d^{2} \cos \left (f x + e\right )^{2} - 15 \, c^{2} + 10 \, c d - 7 \, d^{2} + 4 \,{\left (5 \, c d + 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} \sqrt{d \sin \left (f x + e\right ) + c}}{15 \,{\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{4} - 3 \,{\left (c^{4} d^{2} + 3 \, c^{3} d^{3} + 3 \, c^{2} d^{4} + c d^{5}\right )} f \cos \left (f x + e\right )^{3} -{\left (3 \, c^{5} d + 12 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 18 \, c^{2} d^{4} + 9 \, c d^{5} + 2 \, d^{6}\right )} f \cos \left (f x + e\right )^{2} +{\left (c^{6} + 3 \, c^{5} d + 6 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 9 \, c^{2} d^{4} + 3 \, c d^{5}\right )} f \cos \left (f x + e\right ) +{\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f -{\left ({\left (c^{3} d^{3} + 3 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{3} +{\left (3 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 12 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right )^{2} -{\left (3 \, c^{5} d + 9 \, c^{4} d^{2} + 10 \, c^{3} d^{3} + 6 \, c^{2} d^{4} + 3 \, c d^{5} + d^{6}\right )} f \cos \left (f x + e\right ) -{\left (c^{6} + 6 \, c^{5} d + 15 \, c^{4} d^{2} + 20 \, c^{3} d^{3} + 15 \, c^{2} d^{4} + 6 \, c d^{5} + d^{6}\right )} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/15*(8*d^2*cos(f*x + e)^3 - 4*(5*c*d - d^2)*cos(f*x + e)^2 - 15*c^2 + 10*c*d - 7*d^2 - (15*c^2 + 10*c*d + 11*
d^2)*cos(f*x + e) - (8*d^2*cos(f*x + e)^2 - 15*c^2 + 10*c*d - 7*d^2 + 4*(5*c*d + d^2)*cos(f*x + e))*sin(f*x +
e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^3*d^3 + 3*c^2*d^4 + 3*c*d^5 + d^6)*f*cos(f*x + e)^4
- 3*(c^4*d^2 + 3*c^3*d^3 + 3*c^2*d^4 + c*d^5)*f*cos(f*x + e)^3 - (3*c^5*d + 12*c^4*d^2 + 20*c^3*d^3 + 18*c^2*d
^4 + 9*c*d^5 + 2*d^6)*f*cos(f*x + e)^2 + (c^6 + 3*c^5*d + 6*c^4*d^2 + 10*c^3*d^3 + 9*c^2*d^4 + 3*c*d^5)*f*cos(
f*x + e) + (c^6 + 6*c^5*d + 15*c^4*d^2 + 20*c^3*d^3 + 15*c^2*d^4 + 6*c*d^5 + d^6)*f - ((c^3*d^3 + 3*c^2*d^4 +
3*c*d^5 + d^6)*f*cos(f*x + e)^3 + (3*c^4*d^2 + 10*c^3*d^3 + 12*c^2*d^4 + 6*c*d^5 + d^6)*f*cos(f*x + e)^2 - (3*
c^5*d + 9*c^4*d^2 + 10*c^3*d^3 + 6*c^2*d^4 + 3*c*d^5 + d^6)*f*cos(f*x + e) - (c^6 + 6*c^5*d + 15*c^4*d^2 + 20*
c^3*d^3 + 15*c^2*d^4 + 6*c*d^5 + d^6)*f)*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sin \left (f x + e\right ) + a}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)/(d*sin(f*x + e) + c)^(7/2), x)