3.387 \(\int \frac{\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=102 \[ -\text{Unintegrable}\left (\frac{\tan (a+b x)}{(c+d x)^2},x\right )+\frac{4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{2 \sin (2 a+2 b x)}{d (c+d x)} \]

[Out]

(4*b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d^2 - (2*Sin[2*a + 2*b*x])/(d*(c + d*x)) - (4*b*Sin[
2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/d^2 - Unintegrable[Tan[a + b*x]/(c + d*x)^2, x]

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Rubi [A]  time = 0.341935, 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 = {} \[ \int \frac{\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(4*b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d^2 - (2*Sin[2*a + 2*b*x])/(d*(c + d*x)) - (4*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{\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx &=\int \left (\frac{3 \cos (a+b x) \sin (a+b x)}{(c+d x)^2}-\frac{\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2}\right ) \, dx\\ &=3 \int \frac{\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx-\int \frac{\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx\\ &=3 \int \frac{\sin (2 a+2 b x)}{2 (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\\ &=\frac{3}{2} \int \frac{\sin (2 a+2 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\\ &=-\frac{3 \sin (2 a+2 b x)}{2 d (c+d x)}+\frac{1}{2} \int \frac{\sin (2 a+2 b x)}{(c+d x)^2} \, dx+\frac{(3 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{2 \sin (2 a+2 b x)}{d (c+d x)}+\frac{b \int \frac{\cos (2 a+2 b x)}{c+d x} \, dx}{d}+\frac{\left (3 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 (3 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{3 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{2 \sin (2 a+2 b x)}{d (c+d x)}-\frac{3 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{\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{4 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{2 \sin (2 a+2 b x)}{d (c+d x)}-\frac{4 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*}

Mathematica [A]  time = 4.25604, size = 0, normalized size = 0. \[ \int \frac{\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.663, size = 0, normalized size = 0. \begin{align*} \int{\frac{\sec \left ( bx+a \right ) \sin \left ( 3\,bx+3\,a \right ) }{ \left ( dx+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\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 ) + 2 \,{\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 )}{d^{2} x + c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((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) + 2*(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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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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