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

Optimal. Leaf size=211 \[ \frac{4 a^3 (13 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 (11 A+7 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (7 A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(4*a^3*(7*A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(13*A + 21*
B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(41*A + 42*B)*Sin[c + d*x]
)/(105*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(11*A
+ 7*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.437065, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4017, 3996, 3787, 3771, 2639, 2641} \[ \frac{2 (11 A+7 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (13 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^3 (7 A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(7*A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(13*A + 21*
B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(41*A + 42*B)*Sin[c + d*x]
)/(105*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(11*A
+ 7*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

Rule 4017

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

Rule 3996

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (11 A+7 B)+\frac{1}{2} a (A+7 B) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (11 A+7 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4}{35} \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{2} a^2 (41 A+42 B)+\frac{1}{2} a^2 (8 A+21 B) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (11 A+7 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{8}{105} \int \frac{-\frac{21}{4} a^3 (7 A+9 B)-\frac{5}{4} a^3 (13 A+21 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (11 A+7 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (2 a^3 (7 A+9 B)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (2 a^3 (13 A+21 B)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (11 A+7 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (2 a^3 (7 A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (2 a^3 (13 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (7 A+9 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^3 (13 A+21 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (11 A+7 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 2.51958, size = 194, normalized size = 0.92 \[ \frac{a^3 e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-56 i (7 A+9 B) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+40 (13 A+21 B) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (5 (107 A+84 B) \sin (c+d x)+42 (3 A+B) \sin (2 (c+d x))+168 i (7 A+9 B)+15 A \sin (3 (c+d x)))\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(40*(13*A + 21*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]
- (56*I)*(7*A + 9*B)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)
*(c + d*x))] + Cos[c + d*x]*((168*I)*(7*A + 9*B) + 5*(107*A + 84*B)*Sin[c + d*x] + 42*(3*A + B)*Sin[2*(c + d*x
)] + 15*A*Sin[3*(c + d*x)])))/(210*d*E^(I*d*x))

________________________________________________________________________________________

Maple [A]  time = 1.741, size = 385, normalized size = 1.8 \begin{align*} -{\frac{4\,{a}^{3}}{105\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 120\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -432\,A-84\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 602\,A+294\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -208\,A-126\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +65\,A\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -147\,A\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +105\,B\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -189\,B\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(120*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+(-432*A-84*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(602*A+294*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)+(-208*A-126*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+65*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+105*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*
c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, B\right )} a^{3} \sec \left (d x + c\right )^{3} + 3 \,{\left (A + B\right )} a^{3} \sec \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \sec \left (d x + c\right ) + A a^{3}}{\sec \left (d x + c\right )^{\frac{7}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*a^3*sec(d*x + c)^4 + (A + 3*B)*a^3*sec(d*x + c)^3 + 3*(A + B)*a^3*sec(d*x + c)^2 + (3*A + B)*a^3*s
ec(d*x + c) + A*a^3)/sec(d*x + c)^(7/2), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )}^{3}}{\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*(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)^3/sec(d*x + c)^(7/2), x)