3.228 \(\int \frac{(a+a \sec (c+d x))^{3/2}}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=76 \[ \frac{2 a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.11753, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3813, 21, 3801, 215} \[ \frac{2 a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^{3/2}}{\sqrt{\sec (c+d x)}} \, dx &=\frac{2 a^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+(2 a) \int \frac{\sqrt{\sec (c+d x)} \left (\frac{a}{2}+\frac{1}{2} a \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+a \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.32885, size = 86, normalized size = 1.13 \[ \frac{2 a^2 \left (\sin (c+d x) \sqrt{-(\sec (c+d x)-1) \sec (c+d x)}+\tan (c+d x) \sin ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.187, size = 174, normalized size = 2.3 \begin{align*}{\frac{a}{2\,d\sin \left ( dx+c \right ) } \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -4\,\cos \left ( dx+c \right ) +4 \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 3.42125, size = 370, normalized size = 4.87 \begin{align*} \frac{\sqrt{2}{\left (\sqrt{2} a \log \left (2 \, \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sqrt{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right ) - \sqrt{2} a \log \left (2 \, \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sqrt{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right ) + \sqrt{2} a \log \left (2 \, \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, \sqrt{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right ) - \sqrt{2} a \log \left (2 \, \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, \sqrt{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right ) + 8 \, a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c
) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 +
2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d
*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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Fricas [B]  time = 1.86648, size = 819, normalized size = 10.78 \begin{align*} \left [\frac{4 \, a \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac{4 \,{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, \frac{2 \, a \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{d \cos \left (d x + c\right ) + d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(4*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (a*cos(d*x + c) + a)*sqrt(
a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x +
c) + d), (2*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (a*cos(d*x + c) + a)*s
qrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x +
c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c) + d)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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