3.186 \(\int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=213 \[ \frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{7 d}+\frac{2 a^4 \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{9 d}+\frac{8 a^4 \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{122 a^4 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{32 a^4 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{7 d}+\frac{152 a^4 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}-\frac{152 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d} \]

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

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Rubi [A]  time = 0.251494, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3791, 3768, 3771, 2639, 2641} \[ \frac{2 a^4 \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{9 d}+\frac{8 a^4 \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{122 a^4 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{32 a^4 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{7 d}+\frac{152 a^4 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}-\frac{152 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (a^4 \sec ^{\frac{3}{2}}(c+d x)+4 a^4 \sec ^{\frac{5}{2}}(c+d x)+6 a^4 \sec ^{\frac{7}{2}}(c+d x)+4 a^4 \sec ^{\frac{9}{2}}(c+d x)+a^4 \sec ^{\frac{11}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^{\frac{3}{2}}(c+d x) \, dx+a^4 \int \sec ^{\frac{11}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^{\frac{9}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^{\frac{7}{2}}(c+d x) \, dx\\ &=\frac{2 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{8 a^4 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{12 a^4 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{8 a^4 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{9} \left (7 a^4\right ) \int \sec ^{\frac{7}{2}}(c+d x) \, dx-a^4 \int \frac{1}{\sqrt{\sec (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 \sec ^{\frac{5}{2}}(c+d x) \, dx+\frac{1}{5} \left (18 a^4\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{46 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{32 a^4 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{122 a^4 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{15} \left (7 a^4\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{21} \left (20 a^4\right ) \int \sqrt{\sec (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{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{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{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{152 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{32 a^4 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{122 a^4 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac{1}{15} \left (7 a^4\right ) \int \frac{1}{\sqrt{\sec (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{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{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{152 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{32 a^4 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{122 a^4 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}-\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{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{152 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{32 a^4 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{122 a^4 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [C]  time = 3.86102, size = 289, normalized size = 1.36 \[ \frac{a^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 \left (\frac{1596 \csc (c) \cos (d x)+\tan (c+d x) \left (35 \sec ^3(c+d x)+180 \sec ^2(c+d x)+427 \sec (c+d x)+720\right )}{\sec ^{\frac{7}{2}}(c+d x)}-\frac{12 i \sqrt{2} e^{-i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \left (133 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+60 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+133 \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{2520 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4,x]

[Out]

(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(((-12*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]
*Cos[c + d*x]^4*(133*(1 + E^((2*I)*(c + d*x))) + 133*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeo
metric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 60*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c +
d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (1596*
Cos[d*x]*Csc[c] + (720 + 427*Sec[c + d*x] + 180*Sec[c + d*x]^2 + 35*Sec[c + d*x]^3)*Tan[c + d*x])/Sec[c + d*x]
^(7/2)))/(2520*d)

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Maple [B]  time = 2.649, size = 492, 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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x)

[Out]

-a^4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/72*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-61/90*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-304/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2
*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1544/105*(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))-152/
15*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-1/7*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-16/7*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} \sec \left (d x + c\right )^{5} + 4 \, a^{4} \sec \left (d x + c\right )^{4} + 6 \, a^{4} \sec \left (d x + c\right )^{3} + 4 \, a^{4} \sec \left (d x + c\right )^{2} + a^{4} \sec \left (d x + c\right )\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((a^4*sec(d*x + c)^5 + 4*a^4*sec(d*x + c)^4 + 6*a^4*sec(d*x + c)^3 + 4*a^4*sec(d*x + c)^2 + a^4*sec(d*
x + c))*sqrt(sec(d*x + c)), 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(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)