3.245 \(\int \frac{(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=178 \[ \frac{64 a^3 (5 A+7 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (5 A+7 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(64*a^3*(5*A + 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(5*A + 7*B)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*a*(5*A + 7*B)*(a + a*Sec[c + d*x])^(3/2)*
Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2
))

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Rubi [A]  time = 0.316518, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {4013, 3809, 3804} \[ \frac{64 a^3 (5 A+7 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (5 A+7 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]

[Out]

(64*a^3*(5*A + 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(5*A + 7*B)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*a*(5*A + 7*B)*(a + a*Sec[c + d*x])^(3/2)*
Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2
))

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

Mathematica [A]  time = 0.551393, size = 91, normalized size = 0.51 \[ \frac{2 a^3 \sin (c+d x) \left ((230 A+301 B) \sec ^3(c+d x)+(115 A+98 B) \sec ^2(c+d x)+3 (20 A+7 B) \sec (c+d x)+15 A\right )}{105 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]

[Out]

(2*a^3*(15*A + 3*(20*A + 7*B)*Sec[c + d*x] + (115*A + 98*B)*Sec[c + d*x]^2 + (230*A + 301*B)*Sec[c + d*x]^3)*S
in[c + d*x])/(105*d*Sec[c + d*x]^(5/2)*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.309, size = 121, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 15\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+60\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+115\,A\cos \left ( dx+c \right ) +98\,B\cos \left ( dx+c \right ) +230\,A+301\,B \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))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x)

[Out]

-2/105/d*a^2*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+60*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+115*A*cos(d*x+c)+98*B*cos(
d*x+c)+230*A+301*B)*(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 = 2.11634, size = 520, normalized size = 2.92 \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))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*(5*sqrt(2)*(315*a^2*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) +
77*a^2*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) + 21*a^2*cos(2/7*arct
an2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*a
rctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*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))) - 21*a^2*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))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)
)) + 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arctan2(sin(7/2*d*x
+ 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) + 28*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2
*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

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Fricas [A]  time = 0.46815, size = 317, normalized size = 1.78 \begin{align*} \frac{2 \,{\left (15 \, A a^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (230 \, A + 301 \, B\right )} a^{2} \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))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*A*a^2*cos(d*x + c)^4 + 3*(20*A + 7*B)*a^2*cos(d*x + c)^3 + (115*A + 98*B)*a^2*cos(d*x + c)^2 + (230*
A + 301*B)*a^2*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(c
os(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))**(5/2)*(A+B*sec(d*x+c))/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 (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{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))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)