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

Optimal. Leaf size=74 \[ \frac{4 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}-\frac{2 \sqrt{a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.145952, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac{4 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}-\frac{2 \sqrt{a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx &=-\frac{2 \sqrt{a+a \sin (c+d x)}}{d e (e \cos (c+d x))^{3/2}}+\frac{2 \int \frac{(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{5/2}} \, dx}{a}\\ &=-\frac{2 \sqrt{a+a \sin (c+d x)}}{d e (e \cos (c+d x))^{3/2}}+\frac{4 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.232978, size = 46, normalized size = 0.62 \[ \frac{2 (2 \sin (c+d x)-1) \sqrt{a (\sin (c+d x)+1)}}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.119, size = 44, normalized size = 0.6 \begin{align*}{\frac{ \left ( 4\,\sin \left ( dx+c \right ) -2 \right ) \cos \left ( dx+c \right ) }{3\,d}\sqrt{a \left ( 1+\sin \left ( dx+c \right ) \right ) } \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x)

[Out]

2/3/d*(2*sin(d*x+c)-1)*(a*(1+sin(d*x+c)))^(1/2)*cos(d*x+c)/(e*cos(d*x+c))^(5/2)

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Maxima [B]  time = 1.58294, size = 278, normalized size = 3.76 \begin{align*} -\frac{2 \,{\left (\sqrt{a} \sqrt{e} - \frac{4 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{4 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{\sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )}{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \,{\left (e^{3} + \frac{2 \, e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/3*(sqrt(a)*sqrt(e) - 4*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) + 4*sqrt(a)*sqrt(e)*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^
2 + 1)^2/((e^3 + 2*e^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + e^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d*(sin(d
*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

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Fricas [A]  time = 2.57561, size = 128, normalized size = 1.73 \begin{align*} \frac{2 \, \sqrt{e \cos \left (d x + c\right )} \sqrt{a \sin \left (d x + c\right ) + a}{\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{3 \, d e^{3} \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*(2*sin(d*x + c) - 1)/(d*e^3*cos(d*x + c)^2)

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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(d*x+c))**(1/2)/(e*cos(d*x+c))**(5/2),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 (d x + c\right ) + a}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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