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

Optimal. Leaf size=162 \[ \frac{2 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}}+\frac{34 a^2 \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{68 a^2 \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{68 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac{136 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d} \]

[Out]

(68*a^2*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (34*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*Sqrt[a + a*
Sec[c + d*x]]) + (2*a^2*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (136*a*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (68*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d)

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Rubi [A]  time = 0.275272, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3814, 21, 3803, 3800, 4001, 3792} \[ \frac{2 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}}+\frac{34 a^2 \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{68 a^2 \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{68 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac{136 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(68*a^2*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (34*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*Sqrt[a + a*
Sec[c + d*x]]) + (2*a^2*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (136*a*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (68*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d)

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3803

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-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 3792

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{9} (2 a) \int \frac{\sec ^4(c+d x) \left (\frac{17 a}{2}+\frac{17}{2} a \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{9} (17 a) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{34 a^2 \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{21} (34 a) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{34 a^2 \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{68 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{68}{105} \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{34 a^2 \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{136 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{68 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{1}{45} (34 a) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{68 a^2 \tan (c+d x)}{45 d \sqrt{a+a \sec (c+d x)}}+\frac{34 a^2 \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{136 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{68 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 0.522539, size = 70, normalized size = 0.43 \[ \frac{2 a^2 \tan (c+d x) \left (35 \sec ^4(c+d x)+85 \sec ^3(c+d x)+102 \sec ^2(c+d x)+136 \sec (c+d x)+272\right )}{315 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(272 + 136*Sec[c + d*x] + 102*Sec[c + d*x]^2 + 85*Sec[c + d*x]^3 + 35*Sec[c + d*x]^4)*Tan[c + d*x])/(31
5*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.166, size = 93, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( 272\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}-136\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-34\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-17\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-50\,\cos \left ( dx+c \right ) -35 \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \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]

-2/315/d*a*(272*cos(d*x+c)^5-136*cos(d*x+c)^4-34*cos(d*x+c)^3-17*cos(d*x+c)^2-50*cos(d*x+c)-35)*(a*(cos(d*x+c)
+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/sin(d*x+c)

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

[Out]

Timed out

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Fricas [A]  time = 1.74658, size = 262, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (272 \, a \cos \left (d x + c\right )^{4} + 136 \, a \cos \left (d x + c\right )^{3} + 102 \, a \cos \left (d x + c\right )^{2} + 85 \, a \cos \left (d x + c\right ) + 35 \, a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\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]

2/315*(272*a*cos(d*x + c)^4 + 136*a*cos(d*x + c)^3 + 102*a*cos(d*x + c)^2 + 85*a*cos(d*x + c) + 35*a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [A]  time = 5.21234, size = 243, normalized size = 1.5 \begin{align*} \frac{4 \,{\left (315 \, \sqrt{2} a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (525 \, \sqrt{2} a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (819 \, \sqrt{2} a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 47 \,{\left (2 \, \sqrt{2} a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, \sqrt{2} a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{315 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} 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]

4/315*(315*sqrt(2)*a^6*sgn(cos(d*x + c)) - (525*sqrt(2)*a^6*sgn(cos(d*x + c)) - (819*sqrt(2)*a^6*sgn(cos(d*x +
c)) + 47*(2*sqrt(2)*a^6*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(2)*a^6*sgn(cos(d*x + c)))*tan(1/2*d
*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2
- a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)