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]

68/105*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+68/45*a^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+34/63*a^2*sec(d*x+c)^
3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*a^2*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-136/315*a*(a+a*
sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 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 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]

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 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 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 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)]

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 = 0.55, 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])])

________________________________________________________________________________________

fricas [A]  time = 1.65, size = 98, normalized size = 0.60 \[ \frac {2 \, {\left (272 \, a \cos \left (d x + c\right )^{4} + 136 \, a \cos \left (d x + c\right )^{3} + 102 \, a \cos \left (d x + c\right )^{2} + 85 \, a \cos \left (d x + c\right ) + 35 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 8.34, size = 180, normalized size = 1.11 \[ \frac {4 \, {\left (315 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (525 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (819 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, {\left (2 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.99, size = 93, normalized size = 0.57 \[ -\frac {2 \left (272 \left (\cos ^{5}\left (d x +c \right )\right )-136 \left (\cos ^{4}\left (d x +c \right )\right )-34 \left (\cos ^{3}\left (d x +c \right )\right )-17 \left (\cos ^{2}\left (d x +c \right )\right )-50 \cos \left (d x +c \right )-35\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

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*(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*(1+cos(d*x+c)
)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/sin(d*x+c)*a

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 6.70, size = 429, normalized size = 2.65 \[ \frac {\left (\frac {a\,32{}\mathrm {i}}{9\,d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,32{}\mathrm {i}}{9\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\left (\frac {a\,80{}\mathrm {i}}{7\,d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,176{}\mathrm {i}}{63\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {a\,48{}\mathrm {i}}{5\,d}+\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,352{}\mathrm {i}}{105\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,544{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,272{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a*32i)/(9*d) - (a*exp(c*1i + d*x*1i)*32i)/(9*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1
/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - (((a*80i)/(7*d) - (a*exp(c*1i + d*x*1i)*176i)/(63
*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*
2i) + 1)^3) + (((a*48i)/(5*d) + (a*exp(c*1i + d*x*1i)*352i)/(105*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a*exp(c*1i + d*x*1i)*(a + a/(ex
p(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*544i)/(315*d*(exp(c*1i + d*x*1i) + 1)) - (a*exp(c*1i + d*x
*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*272i)/(315*d*(exp(c*1i + d*x*1i) + 1)*(exp(
c*2i + d*x*2i) + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sec ^{4}{\left (c + d x \right )}\, dx \]

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

________________________________________________________________________________________