3.271 \(\int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{1+\sec (c+d x)}} \, dx\)

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

[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]])

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Rubi [A]  time = 0.234701, antiderivative size = 134, normalized size of antiderivative = 1., 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 = {3823, 4022, 4013, 3807, 215} \[ \frac{2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{\sec (c+d x)+1}}+\frac{26 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{\sec (c+d x)+1}}-\frac{2 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{\sec (c+d x)+1}}-\frac{\sqrt{2} \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

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 3823

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 4022

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]

Rule 4013

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 3807

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 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{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} \operatorname{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*}

Mathematica [A]  time = 0.289526, size = 122, normalized size = 0.91 \[ \frac{\sin (c+d x) \left (2 \sqrt{1-\sec (c+d x)} \left (13 \sec ^2(c+d x)-\sec (c+d x)+3\right )+15 \sqrt{2} \sec ^{\frac{5}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{15 d \sqrt{-\tan ^2(c+d x)} \sec ^{\frac{3}{2}}(c+d x)} \]

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 = 0.21, size = 126, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) +1}{\cos \left ( dx+c \right ) }}} \left ( 6\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-15\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \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 ) -8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+28\,\cos \left ( dx+c \right ) -26 \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

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)

[Out]

-1/15/d*((cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(6*cos(d*x+c)^3-15*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*
(-2/(cos(d*x+c)+1))^(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]  time = 1.98978, size = 478, normalized size = 3.57 \begin{align*} \text{result too large to display} \end{align*}

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 = 1.97565, size = 482, normalized size = 3.6 \begin{align*} \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{align*}

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(-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(1/sec(d*x+c)**(5/2)/(1+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{1}{\sqrt{\sec \left (d x + c\right ) + 1} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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)