3.710 \(\int \frac{\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\)

Optimal. Leaf size=289 \[ \frac{\left (a^2-2 b^2\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac{a \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{3 a \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \]

[Out]

(3*a*Tan[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(2/3)) - (a*AppellF1[1/2, 1/2, -1/3, 3/2, (1 - Sec[c
+ d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(Sqrt[2]*b*(a^2 - b^2)*d*S
qrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) + ((a^2 - 2*b^2)*AppellF1[1/2, 1/2, 2/3, 3/2, (1 -
Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(2/3)*Tan[c + d*x])/(Sqrt[2]*
b*(a^2 - b^2)*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3))

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Rubi [A]  time = 0.35933, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3836, 4007, 3834, 139, 138} \[ \frac{\left (a^2-2 b^2\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac{a \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{3 a \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(5/3),x]

[Out]

(3*a*Tan[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(2/3)) - (a*AppellF1[1/2, 1/2, -1/3, 3/2, (1 - Sec[c
+ d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(Sqrt[2]*b*(a^2 - b^2)*d*S
qrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) + ((a^2 - 2*b^2)*AppellF1[1/2, 1/2, 2/3, 3/2, (1 -
Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(2/3)*Tan[c + d*x])/(Sqrt[2]*
b*(a^2 - b^2)*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3))

Rule 3836

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*Cot[e + f*x]*(
a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] - Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(b*(m + 1) - a*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1]

Rule 4007

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

Rule 3834

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx &=\frac{3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \int \frac{\sec (c+d x) \left (-\frac{2 b}{3}-\frac{1}{3} a \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac{3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac{a \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}+\frac{\left (a^2-2 b^2\right ) \int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{(a \tan (c+d x)) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{2 b \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (\left (a^2-2 b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} (a+b x)^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=\frac{3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{\left (a \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{2 b \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}}}-\frac{\left (\left (a^2-2 b^2\right ) \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}\\ &=\frac{3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac{a F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt{2} b \left (a^2-b^2\right ) d \sqrt{1+\sec (c+d x)} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\left (a^2-2 b^2\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt{2} b \left (a^2-b^2\right ) d \sqrt{1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}\\ \end{align*}

Mathematica [B]  time = 26.2453, size = 7325, normalized size = 25.35 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(5/3),x]

[Out]

Result too large to show

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( a+b\sec \left ( dx+c \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(5/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*sec(d*x+c))**(5/3),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*sec(c + d*x))**(5/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(5/3), x)