3.15 \(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(3/2)+4*a^2*cos(f*x+e)*ln(1-sin(f*x+e))/c^2/f/(a+a*sin(
f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+2*a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c^2/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.55, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2841, 2739, 2740, 2737, 2667, 31} \[ \frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Rule 31

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

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 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 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 2740

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

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps

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

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Mathematica [A]  time = 1.06, size = 169, normalized size = 1.17 \[ \frac {a \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sin (e+f x) \left (2-16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+7\right )}{2 c^2 f (\sin (e+f x)-1)^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + a \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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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(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.37, size = 223, normalized size = 1.55 \[ \frac {\left (8 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-5 \sin \left (f x +e \right )+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+1\right ) \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{f \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(8*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-cos(f*x+e)^2-4*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-8*
ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-5*sin(f*x+e)+4*ln(2/(cos(f*x+e)+1))+1)*(cos(f*x+e)^2-sin(f*x+e)*cos
(f*x+e)+cos(f*x+e)+2*sin(f*x+e)-2)*(a*(1+sin(f*x+e)))^(3/2)/(cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+cos(f*x+e)-2*s
in(f*x+e)-2)/(-c*(sin(f*x+e)-1))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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