3.47 \(\int \frac {(a+b \tan (e+f x))^2}{c+d x} \, dx\)

Optimal. Leaf size=23 \[ \text {Int}\left (\frac {(a+b \tan (e+f x))^2}{c+d x},x\right ) \]

[Out]

Unintegrable((a+b*tan(f*x+e))^2/(d*x+c),x)

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Rubi [A]  time = 0.05, 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 = {} \[ \int \frac {(a+b \tan (e+f x))^2}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*Tan[e + f*x])^2/(c + d*x),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 20.27, size = 0, normalized size = 0.00 \[ \int \frac {(a+b \tan (e+f x))^2}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{d x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*x + c), x)

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maple [A]  time = 2.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{d x +c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^2/(d*x+c),x)

[Out]

int((a+b*tan(f*x+e))^2/(d*x+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*cos(2*f*x + 2*e)^2*log(d*x + c) + 2*b^2*d*sin(2*f*x + 2*e) + ((a^2 - b^
2)*d*f*x + (a^2 - b^2)*c*f)*log(d*x + c)*sin(2*f*x + 2*e)^2 + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*cos(2*f*
x + 2*e)*log(d*x + c) + (d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(2*f*x + 2*e)^2 + (d^2*f*x + c*d*f)*sin(2*f*x
+ 2*e)^2 + 2*(d^2*f*x + c*d*f)*cos(2*f*x + 2*e))*integrate(2*(2*a*b*d*f*x + 2*a*b*c*f + b^2*d)*sin(2*f*x + 2*e
)/(d^2*f*x^2 + 2*c*d*f*x + c^2*f + (d^2*f*x^2 + 2*c*d*f*x + c^2*f)*cos(2*f*x + 2*e)^2 + (d^2*f*x^2 + 2*c*d*f*x
+ c^2*f)*sin(2*f*x + 2*e)^2 + 2*(d^2*f*x^2 + 2*c*d*f*x + c^2*f)*cos(2*f*x + 2*e)), x) + ((a^2 - b^2)*d*f*x +
(a^2 - b^2)*c*f)*log(d*x + c))/(d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(2*f*x + 2*e)^2 + (d^2*f*x + c*d*f)*sin
(2*f*x + 2*e)^2 + 2*(d^2*f*x + c*d*f)*cos(2*f*x + 2*e))

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mupad [A]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^2/(c + d*x),x)

[Out]

int((a + b*tan(e + f*x))^2/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**2/(d*x+c),x)

[Out]

Integral((a + b*tan(e + f*x))**2/(c + d*x), x)

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