3.5.76 \(\int \frac {A+B \sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\) [476]

Optimal. Leaf size=126 \[ \frac {\sqrt {2} B F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}+A \text {Int}\left (\frac {1}{\sqrt [3]{a+b \sec (c+d x)}},x\right ) \]

[Out]

B*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)*2^(1/2)*t
an(d*x+c)/d/(a+b*sec(d*x+c))^(1/3)/(1+sec(d*x+c))^(1/2)+A*Unintegrable(1/(a+b*sec(d*x+c))^(1/3),x)

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Rubi [A]
time = 0.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {A+B \sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(A + B*Sec[c + d*x])/(a + b*Sec[c + d*x])^(1/3),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])/(a + b))^(1/3)*Tan[c + d*x])/(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+B \sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx &=A \int \frac {1}{\sqrt [3]{a+b \sec (c+d x)}} \, dx+B \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\\ &=A \int \frac {1}{\sqrt [3]{a+b \sec (c+d x)}} \, dx-\frac {(B \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{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 {\left (B \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 )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\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]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{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*}

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Mathematica [A]
time = 12.53, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B \sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 0, normalized size = 0.00 \[\int \frac {A +B \sec \left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/3),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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/3),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 = 0.00, 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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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