3.1070 \(\int (a+b \sec (c+d x))^{2/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=247 \[ A \text{Unintegrable}\left ((a+b \sec (c+d x))^{2/3},x\right )+\frac{\sqrt{2} (b B-a C) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{\sqrt{2} C (a+b) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{5}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}} \]

[Out]

(Sqrt[2]*(a + b)*C*AppellF1[1/2, 1/2, -5/3, 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]
*(b*B - a*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)) + A*Unintegrable
[(a + b*Sec[c + d*x])^(2/3), x]

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Rubi [A]  time = 0.322853, 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 = {} \[ \int (a+b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(Sqrt[2]*(a + b)*C*AppellF1[1/2, 1/2, -5/3, 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]
*(b*B - a*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)) + A*Defer[Int][(
a + b*Sec[c + d*x])^(2/3), x]

Rubi steps

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

Mathematica [A]  time = 52.5458, size = 0, normalized size = 0. \[ \int (a+b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.168, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(2/3), 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))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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 )^{\frac{2}{3}} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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