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

Optimal. Leaf size=161 \[ \frac{8 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{7 d}+\frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{12 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

(12*a^2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a^2*Sqrt[Cos[c + d*x]]*Ell
ipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (2*a^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (4*a^2*Sin[
c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^2*Sin[c + d*x])/(7*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.114728, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3788, 3769, 3771, 2639, 4045, 2641} \[ \frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{8 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{12 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a^2*Sqrt[Cos[c + d*x]]*Ell
ipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (2*a^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (4*a^2*Sin[
c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^2*Sin[c + d*x])/(7*d*Sqrt[Sec[c + d*x]])

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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+a \sec (c+d x))^2}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\left (2 a^2\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx+\int \frac{a^2+a^2 \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (6 a^2\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{7} \left (12 a^2\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{1}{7} \left (4 a^2\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (6 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{12 a^2 \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 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{1}{7} \left (4 a^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{12 a^2 \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{8 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{7 d}+\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.78498, size = 149, normalized size = 0.93 \[ \frac{a^2 \left (\frac{672 i \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}+2 \left (-80 i \sqrt{1+e^{2 i (c+d x)}} \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+85 \sin (c+d x)+28 \sin (2 (c+d x))+5 \sin (3 (c+d x))-168 i\right )\right )}{140 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(((672*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 2*(-16
8*I - (80*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]*Sec[c + d*x]
+ 85*Sin[c + d*x] + 28*Sin[2*(c + d*x)] + 5*Sin[3*(c + d*x)])))/(140*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 1.2, size = 272, normalized size = 1.7 \begin{align*} -{\frac{4\,{a}^{2}}{35\,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 ( 40\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-116\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +126\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +10\,{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}}-21\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}}-39\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/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))^2/sec(d*x+c)^(7/2),x)

[Out]

-4/35*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
116*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+126*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+10*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-21*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-39*sin(1/2*d*x+1/2*c)^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)/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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{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))^2/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

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

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

[Out]

integral((a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2)/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+a*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 (a \sec \left (d x + c\right ) + a\right )}^{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))^2/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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