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

Optimal. Leaf size=175 \[ \frac{16 a^2 (15 A+13 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac{2 a (15 A+13 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

[Out]

(64*a^3*(15*A + 13*B)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(15*A + 13*B)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (2*a*(15*A + 13*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d) + (2*(9*A
- 2*B)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*B*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d)

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Rubi [A]  time = 0.352722, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4010, 4001, 3793, 3792} \[ \frac{16 a^2 (15 A+13 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac{2 a (15 A+13 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(15*A + 13*B)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(15*A + 13*B)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (2*a*(15*A + 13*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d) + (2*(9*A
- 2*B)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*B*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d)

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]

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 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 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]

Rubi steps

\begin{align*} \int \sec ^2(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))^{7/2} \tan (c+d x)}{9 a d}+\frac{2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{7 a B}{2}+\frac{1}{2} a (9 A-2 B) \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{21} (15 A+13 B) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{105} (8 a (15 A+13 B)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 (15 A+13 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac{1}{315} \left (32 a^2 (15 A+13 B)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (15 A+13 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac{2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end{align*}

Mathematica [A]  time = 0.642996, size = 96, normalized size = 0.55 \[ \frac{2 a^3 \tan (c+d x) \left (5 (9 A+26 B) \sec ^3(c+d x)+3 (60 A+73 B) \sec ^2(c+d x)+(345 A+292 B) \sec (c+d x)+690 A+35 B \sec ^4(c+d x)+584 B\right )}{315 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(690*A + 584*B + (345*A + 292*B)*Sec[c + d*x] + 3*(60*A + 73*B)*Sec[c + d*x]^2 + 5*(9*A + 26*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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Maple [A]  time = 0.254, size = 141, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 690\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+584\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+345\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+292\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+180\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+219\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,A\cos \left ( dx+c \right ) +130\,B\cos \left ( dx+c \right ) +35\,B \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(690*A*cos(d*x+c)^4+584*B*cos(d*x+c)^4+345*A*cos(d*x+c)^3+292*B*cos(d*x+c)^3+180*
A*cos(d*x+c)^2+219*B*cos(d*x+c)^2+45*A*cos(d*x+c)+130*B*cos(d*x+c)+35*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/c
os(d*x+c)^4/sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 0.486768, size = 346, normalized size = 1.98 \begin{align*} \frac{2 \,{\left (2 \,{\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} +{\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (60 \, A + 73 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, A + 26 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, 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 )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(2*(345*A + 292*B)*a^2*cos(d*x + c)^4 + (345*A + 292*B)*a^2*cos(d*x + c)^3 + 3*(60*A + 73*B)*a^2*cos(d*x
+ c)^2 + 5*(9*A + 26*B)*a^2*cos(d*x + c) + 35*B*a^2)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 5.28357, size = 362, normalized size = 2.07 \begin{align*} \frac{8 \,{\left (315 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (840 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 630 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (945 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (135 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 117 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (15 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, \sqrt{2} B a^{7} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/315*(315*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 315*sqrt(2)*B*a^7*sgn(cos(d*x + c)) - (840*sqrt(2)*A*a^7*sgn(cos(
d*x + c)) + 630*sqrt(2)*B*a^7*sgn(cos(d*x + c)) - (945*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 819*sqrt(2)*B*a^7*sgn
(cos(d*x + c)) - 4*(135*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 117*sqrt(2)*B*a^7*sgn(cos(d*x + c)) - 2*(15*sqrt(2)*
A*a^7*sgn(cos(d*x + c)) + 13*sqrt(2)*B*a^7*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
*tan(1/2*d*x + 1/2*c)^2 + a)*d)