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

Optimal. Leaf size=85 \[ \frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {\sec (c+d x)+1}}+\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d}-\frac {\sinh ^{-1}\left (\frac {\tan (c+d x)}{\sqrt {\sec (c+d x)+1}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3822, 4023, 3807, 215, 3801} \[ \frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {\sec (c+d x)+1}}+\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d}-\frac {\sinh ^{-1}\left (\frac {\tan (c+d x)}{\sqrt {\sec (c+d x)+1}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

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]

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 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 3822

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*d^2
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(2*n - 3)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[d^2/(b*(2*n - 3)), I
nt[((d*Csc[e + f*x])^(n - 2)*(2*b*(n - 2) - a*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] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4023

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

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 111, normalized size = 1.31 \[ \frac {\tan (c+d x) \left (\sqrt {-((\sec (c+d x)-1) \sec (c+d x))}+\sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )+2 \sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right )}{d \sqrt {-\tan ^2(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.97, size = 297, normalized size = 3.49 \[ \frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + 2 \, \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}{\cos \left (d x + c\right ) + 1}\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - 2 \, \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}{\cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(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)) + (cos(d*x + c) +
1)*log(-(cos(d*x + c)^2 + 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)/(cos(d*x + c) + 1)) - (cos(d*x + c) + 1)*log(-(cos(d*x + c)^2 - 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)/(cos(d*x + c) + 1)) + 4*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right ) + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.42, size = 220, normalized size = 2.59 \[ \frac {\left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1+\cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (\arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \cos \left (d x +c \right )-\arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \cos \left (d x +c \right )+4 \cos \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )+2 \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{4 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*(1/cos(d*x+c))^(5/2)*cos(d*x+c)^2*((1+cos(d*x+c))/cos(d*x+c))^(1/2)*(arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2
)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)-arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin
(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)+4*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))+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)^2*(cos(d*x+c)^2-1)

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maxima [B]  time = 1.25, size = 873, normalized size = 10.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(
d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*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*s
qrt(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) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), c
os(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), c
os(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(si
n(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*arct
an2(sin(d*x + c), cos(d*x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*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)
- 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + 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) + 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*co
s(2*d*x + 2*c) + sqrt(2))*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), c
os(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))
*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x
+ c), cos(d*x + c))))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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