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

Optimal. Leaf size=193 \[ \frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.40755, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2739, 2737, 2667, 31} \[ \frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^2}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a^3 \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx}{c^3}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^4 \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^4 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.39388, size = 232, normalized size = 1.2 \[ \frac{a^3 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-3 \sin (3 (e+f x)) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-30 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+18 \cos (2 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+1\right )+9 \sin (e+f x) \left (5 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+4\right )-34\right )}{6 c^3 f (\sin (e+f x)-1)^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(-34 - 30*Log[Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2]] + 18*Cos[2*(e + f*x)]*(1 + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]) + 9*(4 + 5*Log[Cos[(e + f*x)/2
] - Sin[(e + f*x)/2]])*Sin[e + f*x] - 3*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[3*(e + f*x)]))/(6*c^3*f*(
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.168, size = 748, normalized size = 3.9 \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))^(7/2)/(c-c*sin(f*x+e))^(7/2),x)

[Out]

-1/3/f*(-20+3*sin(f*x+e)*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-6*sin(f*x+e)*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))+24*sin(f*x+e)*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-12*ln(2/(cos(f*x+e)+1)
)*sin(f*x+e)*cos(f*x+e)+20*sin(f*x+e)+6*cos(f*x+e)+48*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*ln(2/(cos(
f*x+e)+1))-14*sin(f*x+e)*cos(f*x+e)+28*cos(f*x+e)^2-3*cos(f*x+e)^4*ln(2/(cos(f*x+e)+1))+6*cos(f*x+e)^4*ln(-(-1
+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-12*sin(f*x+e)*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-6*cos(f*x+e)^3+8*sin(f*x+e
)*cos(f*x+e)^3+24*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-48*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))
+24*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)-14*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^4
+18*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-9*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))+24*sin(f*x+e)*
ln(2/(cos(f*x+e)+1))-48*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+12*cos(f*x+e)*ln(2/(cos(f*x+e)+1
))-24*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e)))*(a*(1+sin(f*x+e)))^(7/2)/(sin(f*x+e)*cos(f*x+e)^3
+cos(f*x+e)^4-4*cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^3-4*sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e)^2+8*sin(f*x+e)-4*c
os(f*x+e)+8)/(-c*(-1+sin(f*x+e)))^(7/2)

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Maxima [A]  time = 1.80198, size = 454, normalized size = 2.35 \begin{align*} -\frac{\frac{6 \, a^{\frac{7}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac{7}{2}}} - \frac{3 \, a^{\frac{7}{2}} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac{7}{2}}} + \frac{4 \,{\left (\frac{3 \, a^{\frac{7}{2}} \sqrt{c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{6 \, a^{\frac{7}{2}} \sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{22 \, a^{\frac{7}{2}} \sqrt{c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{6 \, a^{\frac{7}{2}} \sqrt{c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{3 \, a^{\frac{7}{2}} \sqrt{c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{c^{4} - \frac{6 \, c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{15 \, c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{20 \, c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{15 \, c^{4} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{6 \, c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*a^(7/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(7/2) - 3*a^(7/2)*log(sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 1)/c^(7/2) + 4*(3*a^(7/2)*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) - 6*a^(7/2)*sqrt(c)*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 22*a^(7/2)*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^(7/2)*sqrt(c)*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + 3*a^(7/2)*sqrt(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^4 - 6*c^4*sin(f*x + e)/(co
s(f*x + e) + 1) + 15*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 20*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15
*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 6*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^4*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6))/f

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{c^{4} \cos \left (f x + e\right )^{4} - 8 \, c^{4} \cos \left (f x + e\right )^{2} + 8 \, c^{4} + 4 \,{\left (c^{4} \cos \left (f x + e\right )^{2} - 2 \, c^{4}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*
sqrt(-c*sin(f*x + e) + c)/(c^4*cos(f*x + e)^4 - 8*c^4*cos(f*x + e)^2 + 8*c^4 + 4*(c^4*cos(f*x + e)^2 - 2*c^4)*
sin(f*x + e)), x)

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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))**(7/2)/(c-c*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{7}{2}}}{{\left (-c \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))^(7/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

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