3.128 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac {8 a^2 (21 A+19 B) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 a (21 A+19 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d} \]

[Out]

2/35*(7*A-2*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*B*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/a/d+8/105*a^2*(21*A
+19*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*a*(21*A+19*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.30, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4010, 4001, 3793, 3792} \[ \frac {8 a^2 (21 A+19 B) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 a (21 A+19 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(21*A + 19*B)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(21*A + 19*B)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*(7*A - 2*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*B*(a + a*Sec[c
+ d*x])^(5/2)*Tan[c + d*x])/(7*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 3793

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

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]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac {2 \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {5 a B}{2}+\frac {1}{2} a (7 A-2 B) \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 (7 A-2 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac {1}{35} (21 A+19 B) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (21 A+19 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A-2 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac {1}{105} (4 a (21 A+19 B)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {8 a^2 (21 A+19 B) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (21 A+19 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A-2 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 82, normalized size = 0.59 \[ \frac {2 a^2 \tan (c+d x) \left (3 (7 A+13 B) \sec ^2(c+d x)+(63 A+52 B) \sec (c+d x)+2 (63 A+52 B)+15 B \sec ^3(c+d x)\right )}{105 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(2*(63*A + 52*B) + (63*A + 52*B)*Sec[c + d*x] + 3*(7*A + 13*B)*Sec[c + d*x]^2 + 15*B*Sec[c + d*x]^3)*Ta
n[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.42, size = 108, normalized size = 0.78 \[ \frac {2 \, {\left (2 \, {\left (63 \, A + 52 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (63 \, A + 52 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 13 \, B\right )} a \cos \left (d x + c\right ) + 15 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(2*(63*A + 52*B)*a*cos(d*x + c)^3 + (63*A + 52*B)*a*cos(d*x + c)^2 + 3*(7*A + 13*B)*a*cos(d*x + c) + 15*
B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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giac [A]  time = 1.55, size = 215, normalized size = 1.56 \[ \frac {4 \, {\left ({\left ({\left (2 \, \sqrt {2} {\left (21 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (21 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 \, \sqrt {2} {\left (3 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/105*(((2*sqrt(2)*(21*A*a^5*sgn(cos(d*x + c)) + 19*B*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 7*sqrt(2
)*(21*A*a^5*sgn(cos(d*x + c)) + 19*B*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 70*sqrt(2)*(3*A*a^5*sgn(
cos(d*x + c)) + 2*B*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt(2)*(A*a^5*sgn(cos(d*x + c)) + B*
a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a)*d)

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maple [A]  time = 1.43, size = 117, normalized size = 0.85 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (126 A \left (\cos ^{3}\left (d x +c \right )\right )+104 B \left (\cos ^{3}\left (d x +c \right )\right )+63 A \left (\cos ^{2}\left (d x +c \right )\right )+52 B \left (\cos ^{2}\left (d x +c \right )\right )+21 A \cos \left (d x +c \right )+39 B \cos \left (d x +c \right )+15 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*(-1+cos(d*x+c))*(126*A*cos(d*x+c)^3+104*B*cos(d*x+c)^3+63*A*cos(d*x+c)^2+52*B*cos(d*x+c)^2+21*A*cos(d
*x+c)+39*B*cos(d*x+c)+15*B)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)*a

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

[Out]

Timed out

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mupad [B]  time = 6.89, size = 479, normalized size = 3.47 \[ -\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\,\left (7\,A+13\,B\right )\,8{}\mathrm {i}}{105\,d}-\frac {A\,a\,4{}\mathrm {i}}{3\,d}\right )-\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{3\,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 )}+\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\,\left (2\,A+3\,B\right )\,8{}\mathrm {i}}{7\,d}+\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{7\,d}+\frac {A\,a\,4{}\mathrm {i}}{7\,d}\right )-\frac {a\,\left (2\,A+3\,B\right )\,8{}\mathrm {i}}{7\,d}+\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{7\,d}+\frac {A\,a\,4{}\mathrm {i}}{7\,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 ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {A\,a\,4{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+2\,B\right )\,12{}\mathrm {i}}{5\,d}+\frac {B\,a\,16{}\mathrm {i}}{35\,d}\right )-\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+4\,B\right )\,4{}\mathrm {i}}{5\,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 {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}}}\,\left (63\,A+52\,B\right )\,4{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\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))^(3/2))/cos(c + d*x)^2,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*(3*A + 2*B)*4i)/(7*d) -
(a*(2*A + 3*B)*8i)/(7*d) + (A*a*4i)/(7*d)) - (a*(2*A + 3*B)*8i)/(7*d) + (a*(3*A + 2*B)*4i)/(7*d) + (A*a*4i)/(
7*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*(7*A + 13*B)*8i)/(105*d) - (A*a*4i)/(3*d)) - (a*(3*A + 2*B)*4i)/(3*d))
)/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((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*(A + 2*B)*12i)/(5*d) - (A*a*4i)/(5*d) + (B*a*16i)/(35*d)) - (a*(3*A + 2*B)*4i
)/(5*d) + (a*(A + 4*B)*4i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a*exp(c*1i + d*x*1
i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(63*A + 52*B)*4i)/(105*d*(exp(c*1i + d*x*1i)
+ 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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