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

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

[Out]

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

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Rubi [A]  time = 0.129032, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4054, 12, 3793, 3792} \[ \frac{8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (5 B+3 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(5*B + 3*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*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))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{2 \int \frac{1}{2} a (5 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{5 a}\\ &=\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{5} (5 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{15} (4 a (5 B+3 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.253683, size = 62, normalized size = 0.61 \[ \frac{2 a^2 \tan (c+d x) \left ((5 B+9 C) \sec (c+d x)+25 B+3 C \sec ^2(c+d x)+18 C\right )}{15 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(25*B + 18*C + (5*B + 9*C)*Sec[c + d*x] + 3*C*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d
*x])])

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Maple [A]  time = 0.253, size = 95, normalized size = 0.9 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 25\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,B\cos \left ( dx+c \right ) +9\,C\cos \left ( dx+c \right ) +3\,C \right ) }{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/15/d*a*(-1+cos(d*x+c))*(25*B*cos(d*x+c)^2+18*C*cos(d*x+c)^2+5*B*cos(d*x+c)+9*C*cos(d*x+c)+3*C)*(a*(cos(d*x+
c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2/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))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.498846, size = 225, normalized size = 2.23 \begin{align*} \frac{2 \,{\left ({\left (25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*((25*B + 18*C)*a*cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \left (B + C \sec{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 4.52612, size = 238, normalized size = 2.36 \begin{align*} \frac{4 \,{\left (15 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (25 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (5 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt{2} C 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}\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 )}{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(15*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (25*sqrt(2)*B*a^4*sgn(cos(d*x
+ c)) + 15*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - 2*(5*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 3*sqrt(2)*C*a^4*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)/((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)