3.118 \(\int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=187 \[ \frac {2 a (9 A+8 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (9 A+8 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac {8 (9 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {4 a (9 A+8 B) \tan (c+d x)}{45 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

4/105*(9*A+8*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/45*a*(9*A+8*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/6
3*a*(9*A+8*B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*a*B*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+
c))^(1/2)-8/315*(9*A+8*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.34, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4016, 3803, 3800, 4001, 3792} \[ \frac {2 a (9 A+8 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (9 A+8 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac {8 (9 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {4 a (9 A+8 B) \tan (c+d x)}{45 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]

[Out]

(4*a*(9*A + 8*B)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(9*A + 8*B)*Sec[c + d*x]^3*Tan[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*B*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (8*(9*
A + 8*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (4*(9*A + 8*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(105*a*d)

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 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 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]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{9} (9 A+8 B) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (9 A+8 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{21} (2 (9 A+8 B)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (9 A+8 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {4 (9 A+8 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac {(4 (9 A+8 B)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{105 a}\\ &=\frac {2 a (9 A+8 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (9 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {4 (9 A+8 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac {1}{45} (2 (9 A+8 B)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {4 a (9 A+8 B) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (9 A+8 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (9 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {4 (9 A+8 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 98, normalized size = 0.52 \[ \frac {2 a \tan (c+d x) \left (5 (9 A+8 B) \sec ^3(c+d x)+6 (9 A+8 B) \sec ^2(c+d x)+8 (9 A+8 B) \sec (c+d x)+16 (9 A+8 B)+35 B \sec ^4(c+d x)\right )}{315 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]

[Out]

(2*a*(16*(9*A + 8*B) + 8*(9*A + 8*B)*Sec[c + d*x] + 6*(9*A + 8*B)*Sec[c + d*x]^2 + 5*(9*A + 8*B)*Sec[c + d*x]^
3 + 35*B*Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.42, size = 122, normalized size = 0.65 \[ \frac {2 \, {\left (16 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 35 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/315*(16*(9*A + 8*B)*cos(d*x + c)^4 + 8*(9*A + 8*B)*cos(d*x + c)^3 + 6*(9*A + 8*B)*cos(d*x + c)^2 + 5*(9*A +
8*B)*cos(d*x + c) + 35*B)*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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giac [A]  time = 2.28, size = 268, normalized size = 1.43 \[ \frac {2 \, {\left (315 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (630 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (756 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 882 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (522 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 324 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (81 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 107 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

2/315*(315*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 315*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - (630*sqrt(2)*A*a^5*sgn(cos(
d*x + c)) + 420*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - (756*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 882*sqrt(2)*B*a^5*sgn
(cos(d*x + c)) - (522*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 324*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - (81*sqrt(2)*A*a^
5*sgn(cos(d*x + c)) + 107*sqrt(2)*B*a^5*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)^4*sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 1.64, size = 138, normalized size = 0.74 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (144 A \left (\cos ^{4}\left (d x +c \right )\right )+128 B \left (\cos ^{4}\left (d x +c \right )\right )+72 A \left (\cos ^{3}\left (d x +c \right )\right )+64 B \left (\cos ^{3}\left (d x +c \right )\right )+54 A \left (\cos ^{2}\left (d x +c \right )\right )+48 B \left (\cos ^{2}\left (d x +c \right )\right )+45 A \cos \left (d x +c \right )+40 B \cos \left (d x +c \right )+35 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(144*A*cos(d*x+c)^4+128*B*cos(d*x+c)^4+72*A*cos(d*x+c)^3+64*B*cos(d*x+c)^3+54*A*cos(d
*x+c)^2+48*B*cos(d*x+c)^2+45*A*cos(d*x+c)+40*B*cos(d*x+c)+35*B)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)
^4/sin(d*x+c)

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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))^(1/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 10.05, size = 512, normalized size = 2.74 \[ \frac {\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}}\,\left (\frac {A\,16{}\mathrm {i}}{5\,d}+\frac {\left (48\,A-32\,B\right )\,1{}\mathrm {i}}{105\,d}\right )+\frac {\left (336\,A+672\,B\right )\,1{}\mathrm {i}}{105\,d}\right )}{\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 {\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}}\,\left (\frac {A\,16{}\mathrm {i}}{7\,d}-\frac {B\,320{}\mathrm {i}}{63\,d}\right )+\frac {B\,32{}\mathrm {i}}{7\,d}+\frac {\left (144\,A+288\,B\right )\,1{}\mathrm {i}}{63\,d}\right )}{\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 {\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 (-\frac {A\,16{}\mathrm {i}}{9\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,16{}\mathrm {i}}{9\,d}-\frac {\left (16\,A+32\,B\right )\,1{}\mathrm {i}}{9\,d}\right )+\frac {\left (16\,A+32\,B\right )\,1{}\mathrm {i}}{9\,d}\right )}{\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 {{\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}}}\,\left (288\,A+256\,B\right )\,1{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {{\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}}}\,\left (144\,A+128\,B\right )\,1{}\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 + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^4,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*16i)/(5*d) + ((48*A - 3
2*B)*1i)/(105*d)) + ((336*A + 672*B)*1i)/(105*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((a
+ a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*16i)/(7*d) - (B*320i)/(63*
d)) + (B*32i)/(7*d) + ((144*A + 288*B)*1i)/(63*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((
a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*16i)/(9*d) - ((16*A + 32*
B)*1i)/(9*d)) - (A*16i)/(9*d) + ((16*A + 32*B)*1i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^
4) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(288*A + 256*B)*1i)/(31
5*d*(exp(c*1i + d*x*1i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1
/2)*(144*A + 128*B)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))*sec(c + d*x)**4, x)

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