3.107 \(\int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=203 \[ \frac{2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d}+\frac{46 a^3 \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{710 a^3 \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}-\frac{568 a^2 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{693 d}+\frac{284 a^3 \tan (c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{284 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d} \]

[Out]

(284*a^3*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) + (710*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(693*d*Sqrt[a +
a*Sec[c + d*x]]) + (46*a^3*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) - (568*a^2*Sqrt[a + a
*Sec[c + d*x]]*Tan[c + d*x])/(693*d) + (2*a^2*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (
284*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*d)

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Rubi [A]  time = 0.373957, antiderivative size = 203, 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, 4016, 3803, 3800, 4001, 3792} \[ \frac{2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d}+\frac{46 a^3 \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{710 a^3 \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}-\frac{568 a^2 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{693 d}+\frac{284 a^3 \tan (c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{284 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(284*a^3*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) + (710*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(693*d*Sqrt[a +
a*Sec[c + d*x]]) + (46*a^3*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) - (568*a^2*Sqrt[a + a
*Sec[c + d*x]]*Tan[c + d*x])/(693*d) + (2*a^2*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (
284*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*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 4016

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

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))^{5/2} \, dx &=\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{11} (2 a) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{19 a}{2}+\frac{23}{2} a \sec (c+d x)\right ) \, dx\\ &=\frac{46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{99} \left (355 a^2\right ) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{710 a^3 \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{231} \left (710 a^2\right ) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{710 a^3 \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{284 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac{1}{231} (284 a) \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{710 a^3 \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{568 a^2 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{284 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac{1}{99} \left (142 a^2\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{284 a^3 \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{710 a^3 \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{568 a^2 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac{2 a^2 \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{284 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}\\ \end{align*}

Mathematica [A]  time = 0.200337, size = 80, normalized size = 0.39 \[ \frac{2 a^3 \tan (c+d x) \left (63 \sec ^5(c+d x)+224 \sec ^4(c+d x)+355 \sec ^3(c+d x)+426 \sec ^2(c+d x)+568 \sec (c+d x)+1136\right )}{693 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*a^3*(1136 + 568*Sec[c + d*x] + 426*Sec[c + d*x]^2 + 355*Sec[c + d*x]^3 + 224*Sec[c + d*x]^4 + 63*Sec[c + d*
x]^5)*Tan[c + d*x])/(693*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.169, size = 105, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2} \left ( 1136\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}-568\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}-142\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-71\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-131\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-161\,\cos \left ( dx+c \right ) -63 \right ) }{693\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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))^(5/2),x)

[Out]

-2/693/d*a^2*(1136*cos(d*x+c)^6-568*cos(d*x+c)^5-142*cos(d*x+c)^4-71*cos(d*x+c)^3-131*cos(d*x+c)^2-161*cos(d*x
+c)-63)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^5/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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.02099, size = 312, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (1136 \, a^{2} \cos \left (d x + c\right )^{5} + 568 \, a^{2} \cos \left (d x + c\right )^{4} + 426 \, a^{2} \cos \left (d x + c\right )^{3} + 355 \, a^{2} \cos \left (d x + c\right )^{2} + 224 \, a^{2} \cos \left (d x + c\right ) + 63 \, a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\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))^(5/2),x, algorithm="fricas")

[Out]

2/693*(1136*a^2*cos(d*x + c)^5 + 568*a^2*cos(d*x + c)^4 + 426*a^2*cos(d*x + c)^3 + 355*a^2*cos(d*x + c)^2 + 22
4*a^2*cos(d*x + c) + 63*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*
x + c)^5)

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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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.50839, size = 282, normalized size = 1.39 \begin{align*} -\frac{8 \,{\left (693 \, \sqrt{2} a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (1617 \, \sqrt{2} a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3003 \, \sqrt{2} a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 25 \,{\left (99 \, \sqrt{2} a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 4 \,{\left (2 \, \sqrt{2} a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 11 \, \sqrt{2} a^{8} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{693 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{5} \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))^(5/2),x, algorithm="giac")

[Out]

-8/693*(693*sqrt(2)*a^8*sgn(cos(d*x + c)) - (1617*sqrt(2)*a^8*sgn(cos(d*x + c)) - (3003*sqrt(2)*a^8*sgn(cos(d*
x + c)) - 25*(99*sqrt(2)*a^8*sgn(cos(d*x + c)) + 4*(2*sqrt(2)*a^8*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 1
1*sqrt(2)*a^8*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*
d)