3.3.71 \(\int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx\) [271]

Optimal. Leaf size=134 \[ -\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {1+\sec (c+d x)}} \]

[Out]

-arcsinh(tan(d*x+c)/(1+sec(d*x+c)))*2^(1/2)/d+2/5*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(1+sec(d*x+c))^(1/2)-2/15*sin(
d*x+c)/d/sec(d*x+c)^(1/2)/(1+sec(d*x+c))^(1/2)+26/15*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(1+sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3908, 4107, 4098, 3892, 221} \begin {gather*} \frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}+\frac {26 \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {\sec (c+d x)+1}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}-\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]]),x]

[Out]

-((Sqrt[2]*ArcSinh[Tan[c + d*x]/(1 + Sec[c + d*x])])/d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[1 + Se
c[c + d*x]]) - (2*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) + (26*Sqrt[Sec[c + d*x]]*Sin[
c + d*x])/(15*d*Sqrt[1 + Sec[c + d*x]])

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 3892

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

Rule 3908

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

Rule 4098

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

Rule 4107

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

Rubi steps

\begin {align*} \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx &=\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {1}{5} \int \frac {1-4 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2}{15} \int \frac {-\frac {13}{2}+\sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {1+\sec (c+d x)}}-\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {1+\sec (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {1+\sec (c+d x)}}+\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,-\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}\\ &=-\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {1+\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 122, normalized size = 0.91 \begin {gather*} \frac {\left (15 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {5}{2}}(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (3-\sec (c+d x)+13 \sec ^2(c+d x)\right )\right ) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {-\tan ^2(c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]]),x]

[Out]

((15*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(5/2) + 2*Sqrt[1 - Sec[c
+ d*x]]*(3 - Sec[c + d*x] + 13*Sec[c + d*x]^2))*Sin[c + d*x])/(15*d*Sec[c + d*x]^(3/2)*Sqrt[-Tan[c + d*x]^2])

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Maple [A]
time = 0.14, size = 126, normalized size = 0.94

method	result	size
default	\(-\frac {\sqrt {\frac {1+\cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (6 \left (\cos ^{3}\left (d x +c \right )\right )-15 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-8 \left (\cos ^{2}\left (d x +c \right )\right )+28 \cos \left (d x +c \right )-26\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{15 d \sin \left (d x +c \right )}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d*((1+cos(d*x+c))/cos(d*x+c))^(1/2)*(6*cos(d*x+c)^3-15*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*
(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-8*cos(d*x+c)^2+28*cos(d*x+c)-26)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*
x+c)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (114) = 228\).
time = 0.56, size = 354, normalized size = 2.64 \begin {gather*} \frac {\sqrt {2} {\left (60 \, \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 \, \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 ) - 60 \, \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 \, \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 ) - 30 \, \log \left (\cos \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 )^{2} + \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 )^{2} + 2 \, \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 ) + 1\right ) + 30 \, \log \left (\cos \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 )^{2} + \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 )^{2} - 2 \, \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 ) + 1\right ) + 6 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 \, \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 ) + 60 \, \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 )}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/60*sqrt(2)*(60*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*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) - 60*cos(5/2*d*x + 5/2*c)*sin(4/5*a
rctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*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))) - 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 1) + 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(5
/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 1
) + 6*sin(5/2*d*x + 5/2*c) - 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 60*sin(1/5*arcta
n2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/d

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Fricas [A]
time = 2.86, size = 174, normalized size = 1.30 \begin {gather*} \frac {15 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30*(15*(sqrt(2)*cos(d*x + c) + sqrt(2))*log(-(2*sqrt(2)*sqrt((cos(d*x + c) + 1)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c) + cos(d*x + c)^2 - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(3*cos(d*x
+ c)^3 - cos(d*x + c)^2 + 13*cos(d*x + c))*sqrt((cos(d*x + c) + 1)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x +
c)))/(d*cos(d*x + c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(5/2)/(1+sec(d*x+c))**(1/2),x)

[Out]

Integral(1/(sqrt(sec(c + d*x) + 1)*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(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(sec(d*x + c) + 1)*sec(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}+1}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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