3.835 \(\int \frac{1}{\cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=255 \[ \frac{\left (a^2-7 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 b d \left (a^2-b^2\right )^2}+\frac{3 a \left (a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac{3 \left (-2 a^2 b^2+a^4+5 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d (a-b)^2 (a+b)^3}-\frac{a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} (a+b \sec (c+d x))} \]

[Out]

(3*a*(a^2 - 3*b^2)*EllipticE[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) + ((a^2 - 7*b^2)*EllipticF[(c + d*x)/2,
2])/(4*b*(a^2 - b^2)^2*d) + (3*(a^4 - 2*a^2*b^2 + 5*b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*(a - b)
^2*b^2*(a + b)^3*d) - (a^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2) - (3*a^
2*(a^2 - 3*b^2)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.746259, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4264, 3845, 4098, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{\left (a^2-7 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b d \left (a^2-b^2\right )^2}+\frac{3 a \left (a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac{3 \left (-2 a^2 b^2+a^4+5 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d (a-b)^2 (a+b)^3}-\frac{a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

(3*a*(a^2 - 3*b^2)*EllipticE[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) + ((a^2 - 7*b^2)*EllipticF[(c + d*x)/2,
2])/(4*b*(a^2 - b^2)^2*d) + (3*(a^4 - 2*a^2*b^2 + 5*b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*(a - b)
^2*b^2*(a + b)^3*d) - (a^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2) - (3*a^
2*(a^2 - 3*b^2)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x]))

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 3845

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

Rule 4098

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_))^(m_), x_Symbol] :> -Simp[(d*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(
a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(a^2 - b^2)*(m + 1)), x] + Dist[d/(b*(a^2 - b^2)*(m
+ 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) +
b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]
^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

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{1}{\cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{7}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (\frac{a^2}{2}-2 a b \sec (c+d x)-\frac{1}{2} \left (3 a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^2 \left (a^2-3 b^2\right )+a b \left (a^2-4 b^2\right ) \sec (c+d x)+\frac{1}{4} \left (3 a^4-5 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^3 \left (a^2-3 b^2\right )-\left (-a^2 b \left (a^2-4 b^2\right )+\frac{3}{4} a^2 b \left (a^2-3 b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2 b^2 \left (a^2-b^2\right )^2}+\frac{\left (3 \left (a^4-2 a^2 b^2+5 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (3 \left (a^4-2 a^2 b^2+5 b^4\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 b^2 \left (a^2-b^2\right )^2}+\frac{\left (\left (a^2-7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 b \left (a^2-b^2\right )^2}+\frac{\left (3 a \left (a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{3 \left (a^4-2 a^2 b^2+5 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^2 (a+b)^3 d}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (a^2-7 b^2\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b \left (a^2-b^2\right )^2}+\frac{\left (3 a \left (a^2-3 b^2\right )\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{3 a \left (a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2-7 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac{3 \left (a^4-2 a^2 b^2+5 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^2 (a+b)^3 d}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{3 a^2 \left (a^2-3 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.77507, size = 299, normalized size = 1.17 \[ \frac{\frac{\frac{16 b \left (a^2-4 b^2\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac{6 \left (a^2-3 b^2\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-19 a^2 b^2+9 a^4+16 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}-\frac{4 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (3 a \left (a^2-3 b^2\right ) \cos (c+d x)+5 a^2 b-11 b^3\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{16 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

((-4*a^2*Sqrt[Cos[c + d*x]]*(5*a^2*b - 11*b^3 + 3*a*(a^2 - 3*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(
b + a*Cos[c + d*x])^2) + ((2*(9*a^4 - 19*a^2*b^2 + 16*b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)
+ (16*b*(a^2 - 4*b^2)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a +
b) - (6*(a^2 - 3*b^2)*(2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] - 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos
[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b*Sqrt[Si
n[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(16*b^2*d)

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Maple [B]  time = 3.999, size = 1203, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)^2+3/2*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos
(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-3/4/(a+b)/
(a^2-b^2)/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-1/2/(a+b)/(a^2-b^2)/b*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))*a+7/4/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/4*a^3/b^2/(a^2-b^2)^2*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9/4*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c
)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3/4*a^
3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+9/4*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))-3/4/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2
))+3/2/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/4/(a-b)/(
a+b)/(a^2-b^2)*b^2/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*c
os(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(1/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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(1/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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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(1/cos(d*x+c)**(7/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2)), x)