3.475 \(\int \sqrt [3]{a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=125 \[ A \text{Unintegrable}\left (\sqrt [3]{a+b \sec (c+d x)},x\right )+\frac{\sqrt{2} B \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.145576, 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 = {} \[ \int \sqrt [3]{a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 17.8291, size = 0, normalized size = 0. \[ \int \sqrt [3]{a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.139, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a+b\sec \left ( dx+c \right ) } \left ( A+B\sec \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\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+b*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + 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+b*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\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+b*sec(d*x+c))**(1/3)*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(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{\left (B \sec \left (d x + c\right ) + A\right )}{\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+b*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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