3.127 \(\int \frac{\sec ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{11 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{7 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{6 a^2 d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{13 \tan (c+d x)}{3 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(11*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Sec[c + d*x]^2
*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - (13*Tan[c + d*x])/(3*a*d*Sqrt[a + a*Sec[c + d*x]]) + (7*Sqrt
[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

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Rubi [A]  time = 0.288322, antiderivative size = 145, 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 = {3816, 4010, 4001, 3795, 203} \[ \frac{11 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{7 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{6 a^2 d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{13 \tan (c+d x)}{3 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(11*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Sec[c + d*x]^2
*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - (13*Tan[c + d*x])/(3*a*d*Sqrt[a + a*Sec[c + d*x]]) + (7*Sqrt
[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

Rule 3816

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

Rule 4010

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

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, 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 ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^2(c+d x) \left (2 a-\frac{7}{2} a \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{7 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac{\int \frac{\sec (c+d x) \left (-\frac{7 a^2}{4}+\frac{13}{2} a^2 \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{13 \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{7 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac{11 \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{13 \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{7 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac{11 \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 )}{2 a d}\\ &=\frac{11 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{13 \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{7 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.284934, size = 114, normalized size = 0.79 \[ \frac{\tan (c+d x) \left (2 \sqrt{1-\sec (c+d x)} \left (4 \sec ^2(c+d x)-12 \sec (c+d x)-19\right )+33 \sqrt{2} (\sec (c+d x)+1) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{12 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((33*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x]) + 2*Sqrt[1 - Sec[c + d*x]]*(-19 - 12*S
ec[c + d*x] + 4*Sec[c + d*x]^2))*Tan[c + d*x])/(12*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.167, size = 322, normalized size = 2.2 \begin{align*}{\frac{-1+\cos \left ( dx+c \right ) }{24\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) } \left ( 33\,\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+66\,\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +33\,\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -76\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+28\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+64\,\cos \left ( dx+c \right ) -16 \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/24/d/a^2*(-1+cos(d*x+c))*(33*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))^(3/2)*sin(d*x+c)*cos(d*x+c)^2+66*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))^(3/2)*cos(d*x+c)*sin(d*x+c)+33*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))^(3/2)*sin(d*x+c)-7
6*cos(d*x+c)^3+28*cos(d*x+c)^2+64*cos(d*x+c)-16)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 2.20058, size = 1033, normalized size = 7.12 \begin{align*} \left [-\frac{33 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\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 (19 \, \cos \left (d x + c\right )^{2} + 12 \, \cos \left (d x + c\right ) - 4\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, -\frac{33 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\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 (19 \, \cos \left (d x + c\right )^{2} + 12 \, \cos \left (d x + c\right ) - 4\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(33*sqrt(2)*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((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*(19*cos(d*x + c)^2 + 12*cos(d*x + c) - 4)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), -1/12*(33*sqrt(2)
*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + cos(d*x + c))*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*(19*cos(d*x + c)^2 + 12*cos(d*x + c) - 4)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]  time = 10.5126, size = 279, normalized size = 1.92 \begin{align*} -\frac{\frac{{\left ({\left (\frac{3 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{46 \, \sqrt{2}}{\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 )^{2} + \frac{27 \, \sqrt{2}}{\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 )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{33 \, \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 \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/12*(((3*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 46*sqrt(2)/sgn(tan(1/2*d*x + 1/2*c
)^2 - 1))*tan(1/2*d*x + 1/2*c)^2 + 27*sqrt(2)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)/((a*tan(1/
2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 33*sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c)
+ sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d