3.136 \(\int \frac{\sec ^3(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac{\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

(19*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Tan[c + d*x]/(
4*d*(a + a*Sec[c + d*x])^(5/2)) - (13*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.1756, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3799, 4000, 3795, 203} \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac{\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(19*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Tan[c + d*x]/(
4*d*(a + a*Sec[c + d*x])^(5/2)) - (13*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

Rule 3799

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

Rule 4000

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

Rule 3795

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\sec (c+d x) \left (-\frac{5 a}{2}+4 a \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{19 \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{19 \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.626845, size = 116, normalized size = 1.08 \[ \frac{\tan (c+d x) \left (76 \sqrt{2} \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )-2 \sqrt{1-\sec (c+d x)} (13 \sec (c+d x)+9)\right )}{32 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((76*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 - 2*Sqrt[1 - Sec[c + d*
x]]*(9 + 13*Sec[c + d*x]))*Tan[c + d*x])/(32*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.164, size = 323, normalized size = 3. \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{32\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 19\,\sin \left ( dx+c \right ) \ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+38\,\sin \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \cos \left ( dx+c \right ) +19\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +18\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-26\,\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/32/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(19*sin(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2+38*sin(d*x+c)
*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*
x+c))*cos(d*x+c)+19*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+18*cos(d*x+c)^3+8*cos(d*x+c)^2-26*cos(d*x+c))/(cos(d*x+c)+1)/sin(d*x+c)^3

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Maxima [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(sec(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.97066, size = 1062, normalized size = 9.93 \begin{align*} \left [-\frac{19 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{-a} \log \left (\frac{2 \, \sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left (9 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac{19 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) + 2 \,{\left (9 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/
(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(9*cos(d*x + c)^2 + 13*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*
(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*(9*cos(d*x + c)^2 + 13*cos(d*x + c))*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x +
c) + a^3*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]  time = 10.562, size = 216, normalized size = 2.02 \begin{align*} \frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{11 \, \sqrt{2}}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{19 \, \sqrt{2} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 -
1)) + 11*sqrt(2)/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c) + 19*sqrt(2)*log(abs(-sqrt(-a)*ta
n(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d