3.36 \(\int \frac {\sec ^2(a+b x)}{c+d x} \, dx\)

Optimal. Leaf size=19 \[ \text {Int}\left (\frac {\sec ^2(a+b x)}{c+d x},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, 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 = {} \[ \int \frac {\sec ^2(a+b x)}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Sec[a + b*x]^2/(c + d*x),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(a+b x)}{c+d x} \, dx &=\int \frac {\sec ^2(a+b x)}{c+d x} \, dx\\ \end {align*}

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Mathematica [A]  time = 5.62, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^2(a+b x)}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[a + b*x]^2/(c + d*x),x]

[Out]

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

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fricas [A]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (b x + a\right )^{2}}{d x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(b*x + a)^2/(d*x + c), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x + a\right )^{2}}{d x + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}\left (b x +a \right )}{d x +c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (\frac {{\left (b d^{2} x + b c d + {\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \int \frac {\sin \left (2 \, b x + 2 \, a\right )}{{\left (d x + c\right )}^{2} {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )}}\,{d x}}{b} + \sin \left (2 \, b x + 2 \, a\right )\right )}}{b d x + {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + b c + 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b*d^2*x + b*c*d + (b*d^2*x + b*c*d)*cos(2*b*x + 2*a)^2 + (b*d^2*x + b*c*d)*sin(2*b*x + 2*a)^2 + 2*(b*d^2*x
 + b*c*d)*cos(2*b*x + 2*a))*integrate(sin(2*b*x + 2*a)/(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + (b*d^2*x^2 + 2*b*c*d*x
 + b*c^2)*cos(2*b*x + 2*a)^2 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sin(2*b*x + 2*a)^2 + 2*(b*d^2*x^2 + 2*b*c*d*x +
 b*c^2)*cos(2*b*x + 2*a)), x) + sin(2*b*x + 2*a))/(b*d*x + (b*d*x + b*c)*cos(2*b*x + 2*a)^2 + (b*d*x + b*c)*si
n(2*b*x + 2*a)^2 + b*c + 2*(b*d*x + b*c)*cos(2*b*x + 2*a))

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mupad [A]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {1}{{\cos \left (a+b\,x\right )}^2\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^2*(c + d*x)),x)

[Out]

int(1/(cos(a + b*x)^2*(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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