3.2.22 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [122]

Optimal. Leaf size=26 \[ \frac {2 a \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2752} \begin {gather*} \frac {2 a \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(2*a*Sec[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(26)=52\).
time = 0.10, size = 67, normalized size = 2.58 \begin {gather*} \frac {2 (a (1+\sin (c+d x)))^{3/2}}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

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Maple [A]
time = 0.21, size = 37, normalized size = 1.42

method	result	size
default	\(\frac {2 a^{2} \left (1+\sin \left (d x +c \right )\right )}{\cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*a^2*(1+sin(d*x+c))/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. 98 vs. \(2 (24) = 48\).
time = 0.62, size = 98, normalized size = 3.77 \begin {gather*} -\frac {2 \, {\left (a^{\frac {3}{2}} + \frac {2 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{\frac {3}{2}} \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 )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 26, normalized size = 1.00 \begin {gather*} \frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} 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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

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Giac [A]
time = 6.29, size = 38, normalized size = 1.46 \begin {gather*} -\frac {\sqrt {2} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, 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))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.77, size = 37, normalized size = 1.42 \begin {gather*} \frac {4\,a\,\cos \left (c+d\,x\right )\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}}{d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^2,x)

[Out]

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

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