3.3.30 \(\int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [230]

Optimal. Leaf size=116 \[ \frac {8 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sec (c+d x)}}+\frac {2 (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/5*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/5*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c)
)^(1/2)+2/5*a*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3897, 3894, 3889} \begin {gather*} \frac {8 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d \sqrt {a \sec (c+d x)+a}}+\frac {2 \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(5/2),x]

[Out]

(8*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(5*d*Sqrt[Sec[c + d*x]]) + (2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3894

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*m)), x] + Dist[b*((2*m - 1)/(d*m)), Int[(a + b
*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0
] && EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 3897

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[
e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + 1))), x] + Dist[a*(m/(b*d*(m + 1))), Int[(a + b*C
sc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ
[m + n + 1, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {3}{5} \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sec (c+d x)}}+\frac {2 (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} (4 a) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {8 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sec (c+d x)}}+\frac {2 (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 60, normalized size = 0.52 \begin {gather*} \frac {a (13+6 \cos (c+d x)+\cos (2 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{5 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(5/2),x]

[Out]

(a*(13 + 6*Cos[c + d*x] + Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(5*d*Sqrt[Sec[c + d*x
]])

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Maple [A]
time = 0.14, size = 81, normalized size = 0.70

method	result	size
default	\(-\frac {2 \left (\cos ^{3}\left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )-6\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} a}{5 d \sin \left (d x +c \right )}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (98) = 196\).
time = 0.56, size = 210, normalized size = 1.81 \begin {gather*} \frac {\sqrt {2} {\left (20 \, a \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 20 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 2 \, a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 20 \, a \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} \sqrt {a}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/20*sqrt(2)*(20*a*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*a*cos
(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 20*a*cos(5/2*d*x + 5/2*c)*sin
(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*a*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d
*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 2*a*sin(5/2*d*x + 5/2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos
(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*sqrt(a)/d

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Fricas [A]
time = 3.35, size = 80, normalized size = 0.69 \begin {gather*} \frac {2 \, {\left (a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )^{2} + 6 \, a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/5*(a*cos(d*x + c)^3 + 3*a*cos(d*x + c)^2 + 6*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x
+ c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)/sec(d*x+c)**(5/2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)/sec(c + d*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(5/2), x)

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Mupad [B]
time = 1.77, size = 81, normalized size = 0.70 \begin {gather*} \frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (25\,\sin \left (c+d\,x\right )+6\,\sin \left (2\,c+2\,d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}}{10\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(5/2),x)

[Out]

(a*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*(25*sin(c + d*x) + 6*sin(2*c + 2*d*x) + sin(3*c + 3*d*x))*((a*(cos(c +
d*x) + 1))/cos(c + d*x))^(1/2))/(10*d*(cos(c + d*x) + 1))

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