3.3.46 \(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [246]

Optimal. Leaf size=95 \[ \frac {2 \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

2*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)-arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)
*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3906, 3886, 221, 3893, 212} \begin {gather*} \frac {2 \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 221

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]

Rule 3886

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], 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 3893

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*b*(d/
(a*f)), Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(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 3906

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d/b, I
nt[Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]], x], x] - Dist[a*(d/b), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 89, normalized size = 0.94 \begin {gather*} \frac {\left (-2 \text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )+\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*ArcSin[Sqrt[Sec[c + d*x]]] + Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/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] Leaf count of result is larger than twice the leaf count of optimal. \(183\) vs. \(2(78)=156\).
time = 0.14, size = 184, normalized size = 1.94

method result size
default \(\frac {\left (\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-2 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{2 d \sin \left (d x +c \right )^{2} a}\) \(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (78) = 156\).
time = 0.58, size = 476, normalized size = 5.01 \begin {gather*} -\frac {\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 1\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) + 2\right )}{2 \, \sqrt {a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))
)^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x
+ c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1
) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 +
2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))
) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))
^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x
+ c))) + 2) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
 c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos
(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(
d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c)
, cos(d*x + c))) + 2))/(sqrt(a)*d)

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Fricas [A]
time = 3.06, size = 351, normalized size = 3.69 \begin {gather*} \left [\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \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 )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\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 \, a d}, \frac {\sqrt {2} a \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\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 )}{a d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*sqrt(a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 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)))/(a*d), (sqrt(2)*a*sqrt(-1/a)*a
rctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + sqrt(-a)*a
rctan(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)))/(a*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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