3.226 \(\int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=102 \[ \text {Int}\left (\frac {\tan (a+b x)}{(c+d x)^2},x\right )-\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {\sin (2 a+2 b x)}{2 d (c+d x)} \]

[Out]

-b*Ci(2*b*c/d+2*b*x)*cos(2*a-2*b*c/d)/d^2+b*Si(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^2+1/2*sin(2*b*x+2*a)/d/(d*x+c
)+Unintegrable(tan(b*x+a)/(d*x+c)^2,x)

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Rubi [A]  time = 0.18, 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 = {} \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

-((b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d^2) + Sin[2*a + 2*b*x]/(2*d*(c + d*x)) + (b*Sin[2*a
- (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/d^2 + Defer[Int][Tan[a + b*x]/(c + d*x)^2, x]

Rubi steps

\begin {align*} \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx &=-\int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ &=-\int \frac {\sin (2 a+2 b x)}{2 (c+d x)^2} \, dx+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ &=-\left (\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^2} \, dx\right )+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ &=\frac {\sin (2 a+2 b x)}{2 d (c+d x)}-\frac {b \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ &=\frac {\sin (2 a+2 b x)}{2 d (c+d x)}-\frac {\left (b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}+\frac {\left (b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ &=-\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {\sin (2 a+2 b x)}{2 d (c+d x)}+\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 2.40, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sec \left (b x + a\right ) \sin \left (b x + a\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a)^3/(d*x+c)^2,x, algorithm="fricas")

[Out]

integral(-(cos(b*x + a)^2 - 1)*sec(b*x + a)*sin(b*x + a)/(d^2*x^2 + 2*c*d*x + c^2), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.69, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x +a \right ) \left (\sin ^{3}\left (b x +a \right )\right )}{\left (d x +c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (i \, E_{2}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) - i \, E_{2}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 8 \, {\left (d^{2} x + c d\right )} \int \frac {\sin \left (2 \, b x + 2 \, a\right )}{{\left (d x + c\right )}^{2} {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )}}\,{d x} + {\left (E_{2}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) + E_{2}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (d^{2} x + c d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((I*exp_integral_e(2, (2*I*b*d*x + 2*I*b*c)/d) - I*exp_integral_e(2, -(2*I*b*d*x + 2*I*b*c)/d))*cos(-2*(b*
c - a*d)/d) + 8*(d^2*x + c*d)*integrate(sin(2*b*x + 2*a)/(d^2*x^2 + 2*c*d*x + (d^2*x^2 + 2*c*d*x + c^2)*cos(2*
b*x + 2*a)^2 + (d^2*x^2 + 2*c*d*x + c^2)*sin(2*b*x + 2*a)^2 + c^2 + 2*(d^2*x^2 + 2*c*d*x + c^2)*cos(2*b*x + 2*
a)), x) + (exp_integral_e(2, (2*I*b*d*x + 2*I*b*c)/d) + exp_integral_e(2, -(2*I*b*d*x + 2*I*b*c)/d))*sin(-2*(b
*c - a*d)/d))/(d^2*x + c*d)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (a+b\,x\right )}^3}{\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/(cos(a + b*x)*(c + d*x)^2),x)

[Out]

int(sin(a + b*x)^3/(cos(a + b*x)*(c + d*x)^2), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a)**3/(d*x+c)**2,x)

[Out]

Exception raised: HeuristicGCDFailed

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