3.2.33 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [133]

Optimal. Leaf size=55 \[ \frac {8 a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{d} \]

[Out]

-2*a*sec(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d+8*a^2*sec(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.08, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \begin {gather*} \frac {8 a^2 \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{d}-\frac {2 a \sec (c+d x) (a \sin (c+d x)+a)^{3/2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sec[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d - (2*a*Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/d

Rule 2752

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

Rule 2753

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 - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), 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] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{d}+(4 a) \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac {8 a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{d}\\ \end {align*}

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Mathematica [A]
time = 3.13, size = 36, normalized size = 0.65 \begin {gather*} -\frac {2 a^2 \sec (c+d x) (-3+\sin (c+d x)) \sqrt {a (1+\sin (c+d x))}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.34, size = 45, normalized size = 0.82

method	result	size
default	\(-\frac {2 a^{3} \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-3\right )}{\cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (51) = 102\).
time = 0.58, size = 191, normalized size = 3.47 \begin {gather*} -\frac {2 \, {\left (3 \, a^{\frac {5}{2}} - \frac {2 \, a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )}}{d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*a^(5/2) - 2*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 9*a^(5/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*
a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*a^(5/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*a^(5/2)*sin(d*
x + c)^5/(cos(d*x + c) + 1)^5 + 3*a^(5/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)/(d*(sin(d*x + c)/(cos(d*x + c)
+ 1) - 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^(5/2))

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Fricas [A]
time = 0.35, size = 41, normalized size = 0.75 \begin {gather*} -\frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) - 3 \, a^{2}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 6.92, size = 72, normalized size = 1.31 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )} \sqrt {a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(2)*(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + a^2*sgn(cos(-1/4*pi + 1/2
*d*x + 1/2*c))/sin(-1/4*pi + 1/2*d*x + 1/2*c))*sqrt(a)/d

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Mupad [B]
time = 5.46, size = 88, normalized size = 1.60 \begin {gather*} \frac {2\,a^2\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (-22\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+4\,\sin \left (2\,c+2\,d\,x\right )+12\right )}{d\,\left (-4\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*(a*(sin(c + d*x) + 1))^(1/2)*(4*sin(2*c + 2*d*x) - 22*sin(c/2 + (d*x)/2)^2 - 2*sin((3*c)/2 + (3*d*x)/2)
^2 + 12))/(d*(sin(c + d*x) + sin(3*c + 3*d*x) - 4*sin(c + d*x)^2 + 4))

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