3.192 \(\int \frac{(a+a \sec (c+d x))^4}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=187 \[ \frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{7 d}+\frac{122 a^4 \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{32 a^4 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{152 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d} \]

[Out]

(152*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (32*a^4*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (2*a^4*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (8*a^4*S
in[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (122*a^4*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (32*a^4*Sin[c + d*x
])/(7*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.221984, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3791, 3769, 3771, 2639, 2641} \[ \frac{122 a^4 \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{32 a^4 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{152 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(9/2),x]

[Out]

(152*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (32*a^4*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (2*a^4*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (8*a^4*S
in[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (122*a^4*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (32*a^4*Sin[c + d*x
])/(7*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 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))^4}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\int \left (\frac{a^4}{\sec ^{\frac{9}{2}}(c+d x)}+\frac{4 a^4}{\sec ^{\frac{7}{2}}(c+d x)}+\frac{6 a^4}{\sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^4}{\sec ^{\frac{3}{2}}(c+d x)}+\frac{a^4}{\sqrt{\sec (c+d x)}}\right ) \, dx\\ &=a^4 \int \frac{1}{\sec ^{\frac{9}{2}}(c+d x)} \, dx+a^4 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\left (4 a^4\right ) \int \frac{1}{\sec ^{\frac{7}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\left (6 a^4\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^4 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{9} \left (7 a^4\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{3} \left (4 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{7} \left (20 a^4\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{5} \left (18 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 \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 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 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{122 a^4 \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{32 a^4 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{1}{15} \left (7 a^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (20 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{3} \left (4 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 (18 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{46 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{8 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a^4 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{122 a^4 \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{32 a^4 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{1}{15} \left (7 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (20 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{152 a^4 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{32 a^4 \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^4 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{122 a^4 \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{32 a^4 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.25967, size = 156, normalized size = 0.83 \[ \frac{a^4 \left (\frac{51072 i \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}-11520 i \sqrt{1+e^{2 i (c+d x)}} \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+12240 \sin (c+d x)+3556 \sin (2 (c+d x))+720 \sin (3 (c+d x))+70 \sin (4 (c+d x))-25536 i\right )}{2520 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(9/2),x]

[Out]

(a^4*(-25536*I + ((51072*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*
x))] - (11520*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] + 12240*Sin[c + d*x] + 3556*Sin[2*(c + d*x)] + 720*Sin[3*(c + d*x)] + 70*Sin[4*(c + d*x)]))/(2520*d*Sqrt[
Sec[c + d*x]])

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Maple [A]  time = 1.535, size = 260, normalized size = 1.4 \begin{align*} -{\frac{8\,{a}^{4}}{315\,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 ( 280\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}-120\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+34\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+72\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-485\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+180\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -399\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +219\,\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)^(9/2),x)

[Out]

-8/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(280*cos(1/2*d*x+1/2*c)^11-120*cos(1/2*d*x+
1/2*c)^9+34*cos(1/2*d*x+1/2*c)^7+72*cos(1/2*d*x+1/2*c)^5-485*cos(1/2*d*x+1/2*c)^3+180*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-399*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+219*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)^(9/2),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{9}{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)^(9/2),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)^(9/2), 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)**(9/2),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{9}{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)^(9/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(9/2), x)