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

Optimal. Leaf size=101 \[ \frac {8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 A+3 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

2/5*B*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+8/15*a^2*(5*A+3*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a*(5*A+3
*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.14, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4001, 3793, 3792} \[ \frac {8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 A+3 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(5*A + 3*B)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*B*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*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)]

Rubi steps

\begin {align*} \int \sec (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))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+3 B) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (5 A+3 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{15} (4 a (5 A+3 B)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 A+3 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 70, normalized size = 0.69 \[ \frac {2 a \sqrt {a (\sec (c+d x)+1)} ((25 A+18 B) \sin (c+d x)+\tan (c+d x) (5 A+3 B \sec (c+d x)+9 B))}{15 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[a*(1 + Sec[c + d*x])]*((25*A + 18*B)*Sin[c + d*x] + (5*A + 9*B + 3*B*Sec[c + d*x])*Tan[c + d*x]))/(1
5*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.42, size = 89, normalized size = 0.88 \[ \frac {2 \, {\left ({\left (25 \, A + 18 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (d x + c\right ) + 3 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*((25*A + 18*B)*a*cos(d*x + c)^2 + (5*A + 9*B)*a*cos(d*x + c) + 3*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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giac [A]  time = 2.07, size = 170, normalized size = 1.68 \[ \frac {4 \, {\left ({\left (2 \, \sqrt {2} {\left (5 \, A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 5 \, \sqrt {2} {\left (5 \, A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*((2*sqrt(2)*(5*A*a^4*sgn(cos(d*x + c)) + 3*B*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)*(5
*A*a^4*sgn(cos(d*x + c)) + 3*B*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(A*a^4*sgn(cos(d*x
+ c)) + B*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a)*d)

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maple [A]  time = 1.41, size = 95, normalized size = 0.94 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (25 A \left (\cos ^{2}\left (d x +c \right )\right )+18 B \left (\cos ^{2}\left (d x +c \right )\right )+5 A \cos \left (d x +c \right )+9 B \cos \left (d x +c \right )+3 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/d*(-1+cos(d*x+c))*(25*A*cos(d*x+c)^2+18*B*cos(d*x+c)^2+5*A*cos(d*x+c)+9*B*cos(d*x+c)+3*B)*(a*(1+cos(d*x+
c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/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)*(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 = 5.88, size = 213, normalized size = 2.11 \[ -\frac {2\,a\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (A\,25{}\mathrm {i}+B\,18{}\mathrm {i}+A\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,10{}\mathrm {i}+A\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,50{}\mathrm {i}+A\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,10{}\mathrm {i}+A\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,25{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,48{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

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),x)

[Out]

-(2*a*(exp(c*1i + d*x*1i) - 1)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(A*25i + B*18i +
A*exp(c*1i + d*x*1i)*10i + A*exp(c*2i + d*x*2i)*50i + A*exp(c*3i + d*x*3i)*10i + A*exp(c*4i + d*x*4i)*25i + B*
exp(c*1i + d*x*1i)*18i + B*exp(c*2i + d*x*2i)*48i + B*exp(c*3i + d*x*3i)*18i + B*exp(c*4i + d*x*4i)*18i))/(15*
d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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