Optimal. Leaf size=81 \[ \frac {b \text {Ci}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^2}-\frac {\sin ^2(a+b x)}{d (c+d x)}+\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2} \]
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Rubi [A]
time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3394, 12, 3384,
3380, 3383} \begin {gather*} \frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {b \cos \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+b x)}{d (c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3380
Rule 3383
Rule 3384
Rule 3394
Rubi steps
\begin {align*} \int \frac {\sin ^2(a+b x)}{(c+d x)^2} \, dx &=-\frac {\sin ^2(a+b x)}{d (c+d x)}+\frac {(2 b) \int \frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{d}\\ &=-\frac {\sin ^2(a+b x)}{d (c+d x)}+\frac {b \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d}\\ &=-\frac {\sin ^2(a+b x)}{d (c+d x)}+\frac {\left (b \cos \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}+\frac {\left (b \sin \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 {b \text {Ci}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^2}-\frac {\sin ^2(a+b x)}{d (c+d x)}+\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}\\ \end {align*}
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Mathematica [A]
time = 0.28, size = 75, normalized size = 0.93 \begin {gather*} \frac {b \text {Ci}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )-\frac {d \sin ^2(a+b x)}{c+d x}+b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{d^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 156, normalized size = 1.93
method | result | size |
risch | \(-\frac {i b \,{\mathrm e}^{-\frac {2 i \left (d a -c b \right )}{d}} \expIntegral \left (1, 2 i b x +2 i a -\frac {2 i \left (d a -c b \right )}{d}\right )}{2 d^{2}}+\frac {i b \,{\mathrm e}^{\frac {2 i \left (d a -c b \right )}{d}} \expIntegral \left (1, -2 i b x -2 i a -\frac {2 \left (-i a d +i b c \right )}{d}\right )}{2 d^{2}}-\frac {1}{2 d \left (d x +c \right )}+\frac {\left (-2 d x b -2 c b \right ) \cos \left (2 b x +2 a \right )}{4 d \left (-d x b -c b \right ) \left (d x +c \right )}\) | \(155\) |
derivativedivides | \(\frac {-\frac {b^{2}}{2 \left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {b^{2} \left (-\frac {2 \cos \left (2 b x +2 a \right )}{\left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {2 \left (-\frac {2 \sinIntegral \left (-2 b x -2 a -\frac {2 \left (-d a +c b \right )}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}\right )}{4}}{b}\) | \(156\) |
default | \(\frac {-\frac {b^{2}}{2 \left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {b^{2} \left (-\frac {2 \cos \left (2 b x +2 a \right )}{\left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {2 \left (-\frac {2 \sinIntegral \left (-2 b x -2 a -\frac {2 \left (-d a +c b \right )}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}\right )}{4}}{b}\) | \(156\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.36, size = 171, normalized size = 2.11 \begin {gather*} \frac {b^{2} {\left (E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (i \, E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{2}}{4 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 130, normalized size = 1.60 \begin {gather*} \frac {2 \, d \cos \left (b x + a\right )^{2} + 2 \, {\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, d}{2 \, {\left (d^{3} x + c d^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 535 vs.
\(2 (81) = 162\).
time = 2.87, size = 535, normalized size = 6.60 \begin {gather*} \frac {{\left (2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{3} c \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, a b^{2} d \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) - 2 \, b^{3} c \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + 2 \, a b^{2} d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + b^{2} d \cos \left (-\frac {2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}}{d}\right ) - b^{2} d\right )} d^{2}}{2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} d^{4} + b c d^{4} - a d^{5}\right )} b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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