3.7.88 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx\) [688]

Optimal. Leaf size=305 \[ -\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}-\frac {\left (6 a^2-25 b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 a \left (a^2-b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{10 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]

[Out]

-9/40*a*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d+3/8*(a+b*sec(d*x+c))^(5/3)*tan(d*x+c)/b/d-1/40*(6*a^2-25*b^2)*Ap
pellF1(1/2,-2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b^2/d/((a
+b*sec(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)+3/20*a*(a^2-b^2)*AppellF1(1/2,1/3,1/2,3/2,b*(1-sec(d*
x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b*sec(d*x+c))/(a+b))^(1/3)*tan(d*x+c)/b^2/d/(a+b*sec(d*x+c))^(1/3)*2^(1/2)
/(1+sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3925, 4087, 4092, 3919, 144, 143} \begin {gather*} -\frac {\left (6 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{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 )}{20 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} 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 )}{10 \sqrt {2} b^2 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{40 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(40*b*d) + (3*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*b*d)
- ((6*a^2 - 25*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])^(2/3)*Tan[c + d*x])/(20*Sqrt[2]*b^2*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/
3)) + (3*a*(a^2 - b^2)*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])/(a + b))^(1/3)*Tan[c + d*x])/(10*Sqrt[2]*b^2*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(
1/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 3919

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 3925

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4087

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

Rule 4092

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]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx &=\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {3 \int \sec (c+d x) \left (\frac {5 b}{3}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{2/3} \, dx}{8 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {9 \int \frac {\sec (c+d x) \left (\frac {19 a b}{9}-\frac {1}{9} \left (6 a^2-25 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{40 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {1}{40} \left (25-\frac {6 a^2}{b^2}\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx+\frac {\left (3 a \left (a^2-b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{20 b^2}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {\left (\left (-25+\frac {6 a^2}{b^2}\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{40 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (3 a \left (a^2-b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{20 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {\left (\left (-25+\frac {6 a^2}{b^2}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{40 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3}}-\frac {\left (3 a \left (a^2-b^2\right ) \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{20 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac {\left (25-\frac {6 a^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 ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 \sqrt {2} d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 a \left (a^2-b^2\right ) 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]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{10 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(7783\) vs. \(2(305)=610\).
time = 37.79, size = 7783, normalized size = 25.52 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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