3.763 \(\int \frac{A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=239 \[ A \text{Unintegrable}\left (\frac{1}{\sqrt [3]{a+b \sec (c+d x)}},x\right )-\frac{\sqrt{2} a C \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{\sqrt{2} C \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}} \]

[Out]

(Sqrt[2]*C*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])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*a*C*App
ellF1[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])/(b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3)) + A*Unintegrable[(a + b*Sec[c + d
*x])^(-1/3), x]

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Rubi [A]  time = 0.294659, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(1/3),x]

[Out]

(Sqrt[2]*C*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])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*a*C*App
ellF1[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])/(b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3)) + A*Defer[Int][(a + b*Sec[c + d*x
])^(-1/3), x]

Rubi steps

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

Mathematica [A]  time = 76.2612, size = 0, normalized size = 0. \[ \int \frac{A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(1/3),x]

[Out]

Integrate[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(1/3), x]

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Maple [A]  time = 0.163, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2}){\frac{1}{\sqrt [3]{a+b\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(1/3),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(a + b*sec(c + d*x))**(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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