3.264 \(\int \frac{(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=169 \[ \frac{8 a^2 (19 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (19 A+35 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d \sqrt{\sec (c+d x)}}+\frac{6 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(8*a^2*(19*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(19*A + 35*C)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2)) + (6*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

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Rubi [A]  time = 0.413674, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {4087, 4013, 3809, 3804} \[ \frac{8 a^2 (19 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (19 A+35 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d \sqrt{\sec (c+d x)}}+\frac{6 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(8*a^2*(19*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(19*A + 35*C)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2)) + (6*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4013

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

Rule 3809

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*m), x] + Dist[(b*(2*m - 1))/(d*m), Int[(a + b*C
sc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3 a A}{2}+\frac{1}{2} a (2 A+7 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{35} (19 A+35 C) \int \frac{(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (19 A+35 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{105} (4 a (19 A+35 C)) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{8 a^2 (19 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (19 A+35 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.07534, size = 85, normalized size = 0.5 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} ((253 A+140 C) \cos (c+d x)+78 A \cos (2 (c+d x))+15 A \cos (3 (c+d x))+494 A+700 C)}{210 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(a*(494*A + 700*C + (253*A + 140*C)*Cos[c + d*x] + 78*A*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[a*(1 +
Sec[c + d*x])]*Tan[(c + d*x)/2])/(210*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 0.329, size = 108, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 15\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+39\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+52\,A\cos \left ( dx+c \right ) +35\,C\cos \left ( dx+c \right ) +104\,A+175\,C \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{105\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x)

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+39*A*cos(d*x+c)^2+52*A*cos(d*x+c)+35*C*cos(d*x+c)+104*A+175*C)*(
a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)

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Maxima [B]  time = 2.00468, size = 462, normalized size = 2.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*(sqrt(2)*(735*a*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 175*
a*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 63*a*cos(2/7*arctan2(sin
(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 735*a*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(si
n(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*a*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c),
cos(7/2*d*x + 7/2*c))) - 63*a*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))
) + 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(
3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 735*a*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))))*A*sqrt(a) + 280*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a)
)/d

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Fricas [A]  time = 0.486802, size = 292, normalized size = 1.73 \begin{align*} \frac{2 \,{\left (15 \, A a \cos \left (d x + c\right )^{4} + 39 \, A a \cos \left (d x + c\right )^{3} +{\left (52 \, A + 35 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (104 \, A + 175 \, C\right )} a \cos \left (d x + c\right )\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 ) + d\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*A*a*cos(d*x + c)^4 + 39*A*a*cos(d*x + c)^3 + (52*A + 35*C)*a*cos(d*x + c)^2 + (104*A + 175*C)*a*cos(
d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))

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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))**(3/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(7/2), x)