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

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

[Out]

(64*a^3*(7*B + 5*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*B + 5*C)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec
[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

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Rubi [A]  time = 0.174504, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4054, 12, 3793, 3792} \[ \frac{64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (7 B+5 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(7*B + 5*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*B + 5*C)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec
[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 4054

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.35133, size = 79, normalized size = 0.57 \[ \frac{2 a^3 \tan (c+d x) \left (3 (7 B+20 C) \sec ^2(c+d x)+(98 B+115 C) \sec (c+d x)+301 B+15 C \sec ^3(c+d x)+230 C\right )}{105 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.271, size = 119, normalized size = 0.9 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 301\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+230\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+98\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+115\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B\cos \left ( dx+c \right ) +60\,C\cos \left ( dx+c \right ) +15\,C \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/105/d*a^2*(-1+cos(d*x+c))*(301*B*cos(d*x+c)^3+230*C*cos(d*x+c)^3+98*B*cos(d*x+c)^2+115*C*cos(d*x+c)^2+21*B*
cos(d*x+c)+60*C*cos(d*x+c)+15*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^3/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((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.495352, size = 292, normalized size = 2.12 \begin{align*} \frac{2 \,{\left ({\left (301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*((301*B + 230*C)*a^2*cos(d*x + c)^3 + (98*B + 115*C)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c
) + 15*C*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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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((a+a*sec(d*x+c))**(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 4.79268, size = 300, normalized size = 2.17 \begin{align*} -\frac{8 \,{\left (105 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (245 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (49 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (7 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, \sqrt{2} C 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}\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 )}{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/105*(105*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (245*sqrt(2)*B*a^6*sgn(cos
(d*x + c)) + 175*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - 4*(49*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 35*sqrt(2)*C*a^6*sg
n(cos(d*x + c)) - 2*(7*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 5*sqrt(2)*C*a^6*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)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)