3.585 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=189 \[ -\frac{2 a^3 \left (3 c^2+14 c d+43 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}+\frac{2 a^3 (c-d) (3 c+11 d) \cos (e+f x)}{15 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

[Out]

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) + (2*a^3*(c -
d)*(3*c + 11*d)*Cos[e + f*x])/(15*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (2*a
^3*(3*c^2 + 14*c*d + 43*d^2)*Cos[e + f*x])/(15*d^2*(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 = 0.484717, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2762, 2980, 2771} \[ -\frac{2 a^3 \left (3 c^2+14 c d+43 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}+\frac{2 a^3 (c-d) (3 c+11 d) \cos (e+f x)}{15 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) + (2*a^3*(c -
d)*(3*c + 11*d)*Cos[e + f*x])/(15*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (2*a
^3*(3*c^2 + 14*c*d + 43*d^2)*Cos[e + f*x])/(15*d^2*(c + d)^3*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x
]])

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 2980

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]

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{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{7/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac{(2 a) \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-11 d)-\frac{1}{2} a (3 c+7 d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{2 a^3 (c-d) (3 c+11 d) \cos (e+f x)}{15 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{\left (a^2 \left (3 c^2+14 c d+43 d^2\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 d^2 (c+d)^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{2 a^3 (c-d) (3 c+11 d) \cos (e+f x)}{15 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{2 a^3 \left (3 c^2+14 c d+43 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.05218, size = 152, normalized size = 0.8 \[ -\frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (4 \left (7 c^2+46 c d+7 d^2\right ) \sin (e+f x)-\left (3 c^2+14 c d+43 d^2\right ) \cos (2 (e+f x))+89 c^2+42 c d+49 d^2\right )}{15 f (c+d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

-(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(89*c^2 + 42*c*d + 49*d^2 - (3*c^2 + 14
*c*d + 43*d^2)*Cos[2*(e + f*x)] + 4*(7*c^2 + 46*c*d + 7*d^2)*Sin[e + f*x]))/(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 = 0.262, size = 793, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

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))^(7/2),x)

[Out]

-2/15/f/(c+d)^3*(a*(1+sin(f*x+e)))^(5/2)*(c+d*sin(f*x+e))^(1/2)*(439*c*cos(f*x+e)^4*d^4+128*d^5+544*cos(f*x+e)
^2*c^2*d^3+128*c^4*d-256*c^2*d^3+128*c*d^4+523*cos(f*x+e)^4*d^5-432*cos(f*x+e)^2*d^5-262*cos(f*x+e)^6*d^5-128*
c^5*sin(f*x+e)+256*c^2*d^3*sin(f*x+e)-128*c*d^4*sin(f*x+e)-cos(f*x+e)^6*c^2*d^3-181*cos(f*x+e)^6*c*d^4+7*cos(f
*x+e)^4*c^4*d-194*cos(f*x+e)^4*c^3*d^2+16*sin(f*x+e)*cos(f*x+e)^2*c^5-128*d^5*sin(f*x+e)+115*sin(f*x+e)*cos(f*
x+e)^6*d^5+3*cos(f*x+e)^8*c^2*d^3+14*cos(f*x+e)^8*c*d^4-9*cos(f*x+e)^6*c^4*d-27*cos(f*x+e)^6*c^3*d^2-256*c^3*d
^2+48*sin(f*x+e)*cos(f*x+e)^2*c^4*d+79*sin(f*x+e)*cos(f*x+e)^6*c*d^4+50*sin(f*x+e)*cos(f*x+e)^4*c^3*d^2+114*si
n(f*x+e)*cos(f*x+e)^4*c^2*d^3-287*sin(f*x+e)*cos(f*x+e)^4*c*d^4-352*sin(f*x+e)*cos(f*x+e)^2*c^3*d^2-416*sin(f*
x+e)*cos(f*x+e)^2*c^2*d^3+336*sin(f*x+e)*cos(f*x+e)^2*c*d^4+480*c^3*cos(f*x+e)^2*d^2-400*c*d^4*cos(f*x+e)^2-35
5*sin(f*x+e)*cos(f*x+e)^4*d^5+368*sin(f*x+e)*cos(f*x+e)^2*d^5-112*cos(f*x+e)^2*c^4*d-128*sin(f*x+e)*c^4*d+256*
sin(f*x+e)*c^3*d^2-290*cos(f*x+e)^4*c^2*d^3+9*sin(f*x+e)*cos(f*x+e)^6*c^3*d^2+37*sin(f*x+e)*cos(f*x+e)^6*c^2*d
^3+sin(f*x+e)*cos(f*x+e)^4*c^4*d+128*c^5+43*cos(f*x+e)^8*d^5-5*cos(f*x+e)^4*c^5-80*cos(f*x+e)^2*c^5-3*sin(f*x+
e)*cos(f*x+e)^4*c^5)/cos(f*x+e)^5/(cos(f*x+e)^2*d^2+c^2-d^2)^3

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Maxima [B]  time = 2.04848, size = 626, normalized size = 3.31 \begin{align*} -\frac{2 \,{\left ({\left (43 \, c^{3} + 14 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac{5}{2}} - \frac{{\left (15 \, c^{3} - 256 \, c^{2} d - 53 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{{\left (113 \, c^{3} - 116 \, c^{2} d + 493 \, c d^{2} + 50 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \,{\left (17 \, c^{3} - 82 \, c^{2} d + 65 \, c d^{2} - 60 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \,{\left (17 \, c^{3} - 82 \, c^{2} d + 65 \, c d^{2} - 60 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{{\left (113 \, c^{3} - 116 \, c^{2} d + 493 \, c d^{2} + 50 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{{\left (15 \, c^{3} - 256 \, c^{2} d - 53 \, c d^{2} - 6 \, d^{3}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{{\left (43 \, c^{3} + 14 \, c^{2} d + 3 \, c d^{2}\right )} a^{\frac{5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{15 \,{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3} + \frac{{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (c + \frac{2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac{7}{2}} f} \end{align*}

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))^(7/2),x, algorithm="maxima")

[Out]

-2/15*((43*c^3 + 14*c^2*d + 3*c*d^2)*a^(5/2) - (15*c^3 - 256*c^2*d - 53*c*d^2 - 6*d^3)*a^(5/2)*sin(f*x + e)/(c
os(f*x + e) + 1) + (113*c^3 - 116*c^2*d + 493*c*d^2 + 50*d^3)*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*
(17*c^3 - 82*c^2*d + 65*c*d^2 - 60*d^3)*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*(17*c^3 - 82*c^2*d + 6
5*c*d^2 - 60*d^3)*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (113*c^3 - 116*c^2*d + 493*c*d^2 + 50*d^3)*a^(
5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + (15*c^3 - 256*c^2*d - 53*c*d^2 - 6*d^3)*a^(5/2)*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6 - (43*c^3 + 14*c^2*d + 3*c*d^2)*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*(sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 1)/((c^3 + 3*c^2*d + 3*c*d^2 + d^3 + (c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sin(f*x + e)^2/(co
s(f*x + e) + 1)^2)*(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 = 2.08252, size = 1467, normalized size = 7.76 \begin{align*} -\frac{2 \,{\left (32 \, a^{2} c^{2} - 64 \, a^{2} c d + 32 \, a^{2} d^{2} -{\left (3 \, a^{2} c^{2} + 14 \, a^{2} c d + 43 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (11 \, a^{2} c^{2} + 78 \, a^{2} c d - 29 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (23 \, a^{2} c^{2} + 14 \, a^{2} c d + 23 \, a^{2} d^{2}\right )} \cos \left (f x + e\right ) -{\left (32 \, a^{2} c^{2} - 64 \, a^{2} c d + 32 \, a^{2} d^{2} -{\left (3 \, a^{2} c^{2} + 14 \, a^{2} c d + 43 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (7 \, a^{2} c^{2} + 46 \, a^{2} c d + 7 \, a^{2} 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))^(5/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-2/15*(32*a^2*c^2 - 64*a^2*c*d + 32*a^2*d^2 - (3*a^2*c^2 + 14*a^2*c*d + 43*a^2*d^2)*cos(f*x + e)^3 + (11*a^2*c
^2 + 78*a^2*c*d - 29*a^2*d^2)*cos(f*x + e)^2 + 2*(23*a^2*c^2 + 14*a^2*c*d + 23*a^2*d^2)*cos(f*x + e) - (32*a^2
*c^2 - 64*a^2*c*d + 32*a^2*d^2 - (3*a^2*c^2 + 14*a^2*c*d + 43*a^2*d^2)*cos(f*x + e)^2 - 2*(7*a^2*c^2 + 46*a^2*
c*d + 7*a^2*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))**(5/2)/(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{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\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))^(5/2)/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(7/2), x)