3.605 \(\int \frac{(a+b \sec (c+d x))^4}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=211 \[ \frac{2 \left (42 a^2 b^2+5 a^4+21 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^2 \left (5 a^2+39 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{8 a b \left (3 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(8*a*b*(3*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(5*a^4 + 42
*a^2*b^2 + 21*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (36*a^3*b*Sin[c +
d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*a^2*(5*a^2 + 39*b^2)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a^2*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.362749, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3841, 4074, 4047, 3771, 2639, 4045, 2641} \[ \frac{2 a^2 \left (5 a^2+39 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (42 a^2 b^2+5 a^4+21 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{8 a b \left (3 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(7/2),x]

[Out]

(8*a*b*(3*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(5*a^4 + 42
*a^2*b^2 + 21*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (36*a^3*b*Sin[c +
d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*a^2*(5*a^2 + 39*b^2)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a^2*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 4045

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

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+b \sec (c+d x))^4}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+b \sec (c+d x)) \left (9 a^2 b+\frac{1}{2} a \left (5 a^2+21 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (a^2+7 b^2\right ) \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{4}{35} \int \frac{-\frac{5}{4} a^2 \left (5 a^2+39 b^2\right )-7 a b \left (3 a^2+5 b^2\right ) \sec (c+d x)-\frac{5}{4} b^2 \left (a^2+7 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{4}{35} \int \frac{-\frac{5}{4} a^2 \left (5 a^2+39 b^2\right )-\frac{5}{4} b^2 \left (a^2+7 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{5} \left (4 a b \left (3 a^2+5 b^2\right )\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \left (5 a^2+39 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{1}{21} \left (-5 a^4-42 a^2 b^2-21 b^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (4 a b \left (3 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{8 a b \left (3 a^2+5 b^2\right ) \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{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \left (5 a^2+39 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{1}{21} \left (\left (-5 a^4-42 a^2 b^2-21 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{8 a b \left (3 a^2+5 b^2\right ) \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{2 \left (5 a^4+42 a^2 b^2+21 b^4\right ) \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{36 a^3 b \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \left (5 a^2+39 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.799924, size = 142, normalized size = 0.67 \[ \frac{\sqrt{\sec (c+d x)} \left (20 \left (42 a^2 b^2+5 a^4+21 b^4\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+a^2 \sin (2 (c+d x)) \left (15 a^2 \cos (2 (c+d x))+65 a^2+168 a b \cos (c+d x)+420 b^2\right )+336 a b \left (3 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(7/2),x]

[Out]

(Sqrt[Sec[c + d*x]]*(336*a*b*(3*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(5*a^4 + 42*a^2
*b^2 + 21*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + a^2*(65*a^2 + 420*b^2 + 168*a*b*Cos[c + d*x] + 1
5*a^2*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*d)

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Maple [A]  time = 1.425, size = 476, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4/sec(d*x+c)^(7/2),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-360*a^4-672*a^3*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*a^4+672*a^3*b+840*a^2*b^2)*sin(1/2*d*x+1/2
*c)^4*cos(1/2*d*x+1/2*c)+(-80*a^4-168*a^3*b-420*a^2*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*a^4*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+210*a^2*b^2*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+105*b^4*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-252*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-420*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3)/(-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

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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+b*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}{\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+b*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x + c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4)
/sec(d*x + c)^(7/2), x)

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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+b*sec(d*x+c))**4/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 )}^{4}}{\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+b*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4/sec(d*x + c)^(7/2), x)