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

Optimal. Leaf size=161 \[ \frac{136 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{94 a^4 \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{64 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

(64*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (136*a^4*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^4*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (8*a^4*S
in[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (94*a^4*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.187004, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3791, 3769, 3771, 2641, 2639} \[ \frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{94 a^4 \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{136 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{64 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (136*a^4*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^4*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (8*a^4*S
in[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (94*a^4*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^4}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\int \left (\frac{a^4}{\sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a^4}{\sec ^{\frac{5}{2}}(c+d x)}+\frac{6 a^4}{\sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^4}{\sqrt{\sec (c+d x)}}+a^4 \sqrt{\sec (c+d x)}\right ) \, dx\\ &=a^4 \int \frac{1}{\sec ^{\frac{7}{2}}(c+d x)} \, dx+a^4 \int \sqrt{\sec (c+d x)} \, dx+\left (4 a^4\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\left (6 a^4\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^4 \sin (c+d x)}{d \sqrt{\sec (c+d x)}}+\frac{1}{7} \left (5 a^4\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\left (2 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (12 a^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\left (a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\left (4 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{8 a^4 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{94 a^4 \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left (5 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\left (2 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (12 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{64 a^4 \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{6 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{94 a^4 \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left (5 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{64 a^4 \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{136 a^4 \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{2 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{94 a^4 \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.69522, size = 180, normalized size = 1.12 \[ \frac{a^4 \left (\cos \left (\frac{c}{2}\right )-i \sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}\right )+i \sin \left (\frac{c}{2}\right )\right ) \left (\frac{10752 i \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}-2720 i \sqrt{1+e^{2 i (c+d x)}} \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+1910 \sin (c+d x)+336 \sin (2 (c+d x))+30 \sin (3 (c+d x))-5376 i\right )}{420 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(-5376*I + ((10752*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -
E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - (2720*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] + 1910*Sin[c + d*x] + 336*Sin[2*(c + d*x)] + 30*Sin[3*(c +
d*x)]))/(420*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 1.352, size = 272, normalized size = 1.7 \begin{align*} -{\frac{8\,{a}^{4}}{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 ( 60\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-258\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +448\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +85\,{\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}}-168\,{\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}}-167\, \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))^4/sec(d*x+c)^(7/2),x)

[Out]

-8/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(60*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-258*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+85*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)-168*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)-167*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(-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))^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{a^{4} \sec \left (d x + c\right )^{4} + 4 \, a^{4} \sec \left (d x + c\right )^{3} + 6 \, a^{4} \sec \left (d x + c\right )^{2} + 4 \, a^{4} \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+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((a^4*sec(d*x + c)^4 + 4*a^4*sec(d*x + c)^3 + 6*a^4*sec(d*x + c)^2 + 4*a^4*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+a*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 (a \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+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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