3.137 \(\int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac {64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (7 A+5 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 A+5 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

2/35*a*(7*A+5*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)/d+64/105*a^3*(7*A
+5*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(7*A+5*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.18, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4001, 3793, 3792} \[ \frac {64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (7 A+5 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 A+5 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(7*A + 5*B)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*A + 5*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*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))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{7} (7 A+5 B) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (7 A+5 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 d}+\frac {1}{35} (8 a (7 A+5 B)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (7 A+5 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 A+5 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 d}+\frac {1}{105} \left (32 a^2 (7 A+5 B)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (7 A+5 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 A+5 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 d}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 89, normalized size = 0.64 \[ \frac {2 a^2 \sqrt {a (\sec (c+d x)+1)} \left ((301 A+230 B) \sin (c+d x)+\tan (c+d x) \left (3 (7 A+20 B) \sec (c+d x)+98 A+15 B \sec ^2(c+d x)+115 B\right )\right )}{105 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[a*(1 + Sec[c + d*x])]*((301*A + 230*B)*Sin[c + d*x] + (98*A + 115*B + 3*(7*A + 20*B)*Sec[c + d*x]
+ 15*B*Sec[c + d*x]^2)*Tan[c + d*x]))/(105*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.43, size = 115, normalized size = 0.83 \[ \frac {2 \, {\left ({\left (301 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (98 \, A + 115 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/105*((301*A + 230*B)*a^2*cos(d*x + c)^3 + (98*A + 115*B)*a^2*cos(d*x + c)^2 + 3*(7*A + 20*B)*a^2*cos(d*x + c
) + 15*B*a^2)*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 = 1.78, size = 216, normalized size = 1.57 \[ \frac {8 \, {\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

8/105*((4*(2*sqrt(2)*(7*A*a^6*sgn(cos(d*x + c)) + 5*B*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 7*sqrt(2
)*(7*A*a^6*sgn(cos(d*x + c)) + 5*B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(7*A*a^6*sgn(co
s(d*x + c)) + 5*B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt(2)*(A*a^6*sgn(cos(d*x + c)) + B*a^
6*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 = 1.33, size = 119, normalized size = 0.86 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (301 A \left (\cos ^{3}\left (d x +c \right )\right )+230 B \left (\cos ^{3}\left (d x +c \right )\right )+98 A \left (\cos ^{2}\left (d x +c \right )\right )+115 B \left (\cos ^{2}\left (d x +c \right )\right )+21 A \cos \left (d x +c \right )+60 B \cos \left (d x +c \right )+15 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(301*A*cos(d*x+c)^3+230*B*cos(d*x+c)^3+98*A*cos(d*x+c)^2+115*B*cos(d*x+c)^2+21*A*cos(
d*x+c)+60*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^2

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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))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 6.35, size = 590, normalized size = 4.28 \[ \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\,a^2\,2{}\mathrm {i}}{d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (301\,A+230\,B\right )\,2{}\mathrm {i}}{105\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1}-\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^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,A+10\,B\right )\,2{}\mathrm {i}}{7\,d}\right )+\frac {A\,a^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,A+10\,B\right )\,2{}\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^2\,\left (5\,A+2\,B\right )\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (5\,A+9\,B\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (7\,A-8\,B\right )\,2{}\mathrm {i}}{35\,d}\right )-\frac {A\,a^2\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (A+2\,B\right )\,2{}\mathrm {i}}{d}+\frac {a^2\,\left (A+B\right )\,4{}\mathrm {i}}{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^2\,\left (5\,A+2\,B\right )\,2{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (63\,A+80\,B\right )\,2{}\mathrm {i}}{105\,d}\right )-\frac {A\,a^2\,2{}\mathrm {i}}{3\,d}+\frac {a^2\,\left (9\,A+10\,B\right )\,2{}\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x),x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*2i)/d - (a^2*exp(c*1i + d*x*1i)*(301*A
+ 230*B)*2i)/(105*d)))/(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)*(exp(c*1i + d*x*1i)*((A*a^2*2i)/(7*d) + (a^2*(3*A + 4*B)*10i)/(7*d) - (a^2*(5*A + 2*B)*2i)/(7*d) - (a^2*(11
*A + 10*B)*2i)/(7*d)) + (A*a^2*2i)/(7*d) + (a^2*(3*A + 4*B)*10i)/(7*d) - (a^2*(5*A + 2*B)*2i)/(7*d) - (a^2*(11
*A + 10*B)*2i)/(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^2*(5*A + 2*B)*2i)/(5*d) - (a^2*(5*A + 9*B)*4i)/(5*d) +
(a^2*(7*A - 8*B)*2i)/(35*d)) - (A*a^2*2i)/(5*d) - (a^2*(A + 2*B)*2i)/d + (a^2*(A + B)*4i)/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^2*(5*A + 2*B)*2i)/(3*d) - (a^2*(63*A + 80*B)*2i)/(105*d)) - (A*a^2*2i)/(3*d) + (a^2*(9*A + 1
0*B)*2i)/(3*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 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/2)*(A + B*sec(c + d*x))*sec(c + d*x), x)

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